Mathematics Tricks & Tips of Circle for NDA

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Tricks & Tips of Circle for NDA

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Basic geometrical concepts related to Circle

.

Finding equation of circle under different cases

`(x - a)^2 + (y - b)^2 = r^2` .

`(a, b) =` centre and `r =` radius

Q 2328878701

What is the equation of circle which touches the
lines `x = 0, y = 0` and `x = 2`?
NDA Paper 1 2007

(A)

`x^2 + y^2 + 2x + 2 y + 1 = 0`

(B)

`x^2 + y^2 - 4x- 4y + 1 = 0`

(C)

`x^2 + y^2 -2x- 2y + 1 = 0`

(D)

none of the above

Solution:

From the figure it is clear that coordinates of centre of

circle are `(1, 1)` and radius of circle is `1`.

`:. ` Equation of circle is

`(x-1)^2 + (y-1)^2 =1`

`=> x^2 -2x +1 + y^2 -2y +1 =1`

`=> x^2 + y^2 -2x -2y +1 =0`

Correct Answer is `=>` (C) `x^2 + y^2 -2x- 2y + 1 = 0`

Q 2318067800

The circle `x^2 + y^2 + 4x - 4y + 4 = 0` touches
NDA Paper 1 2009

(A)

Only the `X`-axis

(B)

Only the `Y`-axis

(C)

Both the axes

(D)

Neither of the axes

Solution:

Given equation is

`x^2 + 4x + 4 + y^2 - 4y = 0`

`=> (x+2)^2 +(y-2)^2= 2^2`

Here, we see that the circle touches both the axes.

Correct Answer is `=>` (C) Both the axes

Q 2318178009

The equation of the circle which touches the axes
at a distance `5` from the origin is `y^2 + x^2 - 2 alpha x- 2 alpha y + alpha^2 = 0`. What is the value
of `alpha`?
NDA Paper 1 2008

(A)

`4`

(B)

`5`

(C)

`6`

(D)

`7`

Solution:

Given that the circle touches the axes at a distance `5`

from the origin, then

So, equation of the circle from the figure is
`(x- 5)^2 + (y- 5)^2= (5)^2`

`x^2 + 25- 10x + y^2 + 25- 10y = 25`

`=> x^2 + y^2 - 10x- 10y + 25 = 0`

Comparing with.

`x^2 + y^2 -2 alpha x -2 alpha y+ alpha^2 =0 => alpha =5`

Correct Answer is `=>` (B) `5`

Q 2328878701

What is the equation of circle which touches the
lines `x = 0, y = 0` and `x = 2`?
NDA Paper 1 2007

(A)

`x^2 + y^2 + 2x + 2 y + 1 = 0`

(B)

`x^2 + y^2 - 4x- 4y + 1 = 0`

(C)

`x^2 + y^2 -2x- 2y + 1 = 0`

(D)

none of the above

Correct Answer is `=>` (C) `x^2 + y^2 -2x- 2y + 1 = 0`

Q 2761680525

What is the radius of the circle passing through the point (2, 4) and having centre at the intersection of the lines x -y = 4 and 2x +3y +7 =0 ?
NDA Paper 1 2016

(A)

3 units

(B)

5 units

(C)

`3sqrt3` units

(D)

`5sqrt2 ` units

Solution:

Centre `x-y-4 =0`

`2x+3y +7 =0`

`x=1 , y=3`

Radius of circle will be the distance between centre and passing point `r= sqrt((2-1)^2+(4+3)^2) = 5 sqrt 2`

Correct Answer is `=>` (D) `5sqrt2 ` units

..General Equation Of a Circle

.Expand (1) `x^2 + y^2 - 2ax - 2by + a^2 + b^2 - r^2 = 0`.

or

The equation `ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0`

So this equation represents a circle, if `a = b` and `h = 0`

hence the general equation of the circle is taken as

`x^2+y^2+2gx+2fy+c=0`.................(2)

This can be written as `(x + g)^2 + (y + f)^2 = (sqrt(g^2+f^2 -c))^2`

hence, Centre `= ( - g, - f)`

i.e (`-1/2 ` coefficient of `x;-1/2` coefficient of `y`)

Radius `equiv sqrt(g^2+f^2-c)`

Q 2368467305

For the equation `ax^2 + by^2 + 2hxy + 2gx`

`+ 2 fy + c = 0`, where `a ne 0`, to represent a, circle, the
required condition will be
NDA Paper 1 2011

(A)

`a = b` and `c = 0`

(B)

`f=g` and `h=0`

(C)

`a=b` and `h=0`

(D)

`f=g` and `c=0`

Solution:

The equation `ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0`

represents a circle, if `a = b` and `h = 0`

Then, the equation becomes the general equation of a circle

`x^2 + y^2 + 2gx + 2fy + c = 0`

Correct Answer is `=>` (C) `a=b` and `h=0`

Q 2147178983

Consider a circle passing through the origin and the points `(a, b)`
and `(-b, -a)`.

On which line does the centre of the circle lie?
NDA Paper 1 2016

(A)

`x + y = 0`

(B)

`x - y = 0`

(C)

`x + y = a + b`

(D)

`x - y = a^2 - b^2`

Solution:

Given that circle passes through

`(0, 0), (a, b)` and `(-b,- a)`

Now, equation of circle is

`x^2 + y^2 + 2gx + 2fy + c = 0`

When `x = 0, y = 0 => 0 + 0 + 0 + 0 + c = 0`

`=> c = 0` .......(i)

When `x = a , y = b`

` a^2 + b^2 + 2ga + 2fb = 0 quad [ ∵ c = 0 ]`

` => ga + fb = -1/2 (a^2 + b^2 )` .......(ii)

When `x = - b , y = - a`

` b^2 + a^2 - 2gb - 2fa = 0`

`=> gb + fa = 1/2 (a^2 + b^2)` ... (iii)

On multiplying Eq. (ii) by a and Eq. (iii) by we get

`a^2g + abf = -1/2 (a^2 + b^2 )*a`

`b^2g + abf = 1/2 (a^2 + b^2 )· b`
- - -
_________________

` (a^2 - b^2)g = -1/2 (a^2 + b^2) ( a + b) `

` => g = -1/2 ((a^2 + b^2) (a + b))/((a^2 - b^2) ) `

` => g = -1/2 ((a^2 + b^2)/(a - b))` ............(iv)

Again, on multiplying Eq. (ii) by band Eq. (iii) by a, we
get

` abg + fb^2 = - 1/2 ( a^2 + b^2 ) ·b`

` abg + fa^2 = 1/2 (a^2 + b^2 ) ·a`
- - -
________________

` f(a^2 - b^2) = -1/2 (a^2 - b^2) (b +a)`

` => f = -1/2 (a^2 + b^2)/(b^2 - a^2) ( b + a)`

` f = -1/2 (( a^2 + b^2)/(a - b))` ............(v)

From Eqs. (iv) and (v), we get

`f = -g => y = -x`

`:.` Required equation of line is `x + y = 0`

Correct Answer is `=>` (A) `x + y = 0`

Q 2318856700

The radius of the circle `x^2 + y^2 + x + c = 0`
passing through the origin is
NDA Paper 1 2013

(A)

`1/4`

(B)

`1/2`

(C)

`1`

(D)

`2`

Solution:

Given equation of circle is

`x^2 + y^2 + x + c = 0`.................(i)

Since, the circle passes through the origin.

`:. (0)^2 + (0)^2 + 0 + c = 0`

`=> c = 0`

From Eq. (i),

`x^2 +y^2 +x=0`

`=> x^2+y^2+x +1/4 =1/4`

`=> (x+1/2)^2 + (y-0)^2 = (1/2)^2`

So, the required radius of circle is `1/2`

Correct Answer is `=>` (B) `1/2`

Q 1679167916

Consider the circles `x^2 + y^ 2 + 2ax + c = 0`
and `x^2 + y^ 2 + 2by + c = 0`.

What is the distance between the centres of the two
circles?
NDA Paper 1 2014

(A)

` sqrt( a^2 + b^2)`

(B)

`( a^2 + b^2)`

(C)

` a + b`

(D)

` 2(a + b)`

Solution:

Equations of circles are

`x^2 + y^ 2 + 2ax + c = 0`

and `x^2 + y^ 2 + 2by + c = 0`

Since, the centers of two circles are `(-a, 0)` and `(0, -b)`.

:. Distance between two centres `= sqrt( a^2 + b^2)`

Correct Answer is `=>` (A) ` sqrt( a^2 + b^2)`

Q 2328667501

Consider the. following statements in respect of
circles `x^2 + y^2 -2x -2y =0` and `x^2 + y^2 = 1`

I. The radius of the first circle is twice that of the
second circle.

II. Both the circles pass through the origin.

Which of the statement(s) given above is/are correct?
NDA Paper 1 2010

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

The equation of first circle is `x^2 + y^2 - 2x- 2 y = 0`.

Radius of this circle`= sqrt ((-1)^2 + (-1)^2) = sqrt(2)`

and equation of second circle is `x^2 + y^2 = 1`.

Radius of this circle `= 1`.

From above it is clear that the radius of first circle is not twice that
of the second circle

Also, the first circle passes through the origin while the second
circle does not pass througl1 the origin.

Hence, neither Statements I or II is correct.

Correct Answer is `=>` (D) Neither I nor II

..Diameter Form Of a Circle

.The equation of the circle drawn on the straight line joining two given points `(x_1 ,y_1)` and `(x_2, y_2)` as diameter is

`(x-x_1)(x-x_2)+ (y-y_1) (y-y_2)=0`

Q 2348478303

Equation of a circle passing through origin is
`x^2 + y^2 - 6x + 2y = 0`. What is the equation of
one of its diameters?
NDA Paper 1 2008

(A)

`x+ 3y = 0`

(B)

`x + y =0`

(C)

`x =y`

(D)

`3x + y =0`

Solution:

The given equation of circle is

`x^2 + y^2 - 6x + 2y = 0`

`=> x(x- 6) + y(y + 2) = 0`

`=> (x - 0) (x- 6) + (y- 0)(y + 2) = 0`

This is the equation of circle in diameter form where end points of
the diameter are `(0, 0)` and `(6, -2)`.

Now, the equation of diameter is a line which pa!;ses through the
points `(0, 0)` and `(6, - 2)` which is ·

`(y - 0) = -2/6 (x-0)`

`=> x+3y = 0`

Correct Answer is `=>` (A) `x+ 3y = 0`

Q 2802612538

Find the equation of circle if
(i) Centre is at origin & radius `3`
(ii) Circle passes through origin & centre `(1, 2)`
(iii) Circle touchs x-axis & centre is `(3, 2)`
(iv) Circle touches the both the co-ordinates axes in first quadrant and radius `= 3`
(v) Circle passes through the origin centre lies on positive y-axis at `(0, 3)`
(vi) Circle is concentric with circle `x^2 + y^2 - 8x + 6y - 5 = 0` and passing through the point `(- 2, - 7)`.

Solution:

(i) Centre `(0, 0)` radius `= 3`

`(x - 0)^2 + (y - 0)^2 = 3^2`

`x^2 +y^2 =9`

(ii) Centre `(1, 2)`

radius `= sqrt((1-0)^2 +(2-0)^2) = sqrt 5`

Equation of circle

`(x - 1)^2+(y - 2)^2 = (sqrt 5)^2`

`x^2 + y^2 - 2x - 4y = 0`

(iii) Centre `(3, 2)`

Circle touches `x`-axis

`(x - 3)^2 + (y - 2)^2 = 2^2`

`x^2 + y^2 - 6x - 4y + 9 = 0`

(iv) radius `= 3`

centre `(3, 3)`

Equation of circle

`(x - 3)^2 + (y - 3)^2 = 3^ 2`

`=> x^2 + y^2 - 6x - 6y + 9 = 0`

(v) Centre `(0, 3)`

radius `= 3`

`(x - 0)^2 + (y - 3)^2 = 3^2`

`x^2+y^2-6y =0`

(vi) Centre `(4, - 3)`

& passes through `(-2, -7)`

radius `= sqrt((4+2)^2 +(4)^2) = sqrt(36 +16) = sqrt 52`

Equation of circle

`(x - 4)^2 + (y + 3)^2 = 52`

`x^2 + y^2 - 8x + 6y - 27 = 0`

..Parametric Equation Of a Circle

Let the equation of circle is

`(x - x_1)^2+ (y - y _1)^2 = r^2`, where centre is `(x_1, y _1)` & radius is `'r'` then, by diagram

`x - x_1= r cos theta` and `y - y _1 = r sin theta`

or `x = x_ 1 +r cos theta` and `y = y_1 + r sin theta`

`:. x=x_1 + r cos theta` & `y=y_1+ r sin theta ->` Parametric equation of circle where `theta in [0, 2 pi) ` is any parameter

Q 2569367215

Find the parametric form of the equation of the circle `x^2+y^2+px+py = 0`

Solution:

Equation of the circle can be re-written in the form

`(x+p/2)^2+(y+p/2)^2 = p^2/2`

Therefore, the parametric form of the equation of the given circle is

`x = -p/2+p/sqrt2 costheta = p/2 (-1+sqrt2 cos theta)`

and `y = -p/2+p/sqrt2 sintheta = p/2(-1+sqrt2 sintheta)` where `0 le theta < 2pi`

Q 2802612538

Position of a point with respect to a circle

Let any point P, we've to find its position w.r.t a circle `S = 0`

(i) `S(P) < 0` point is inside the circle
(ii) `S(P) = 0` point is on the circle
(iii) `s(P) > 0` point is onside the circle

Q 2388056807

Which one of the following points lies inside a
circle of radius `6` and centre at `(3, 5)`?
NDA Paper 1 2013

(A)

`{-2, -1)`

(B)

`(0, 1)`

(C)

`(-1,-2)`

(D)

`(2,-1)`

Solution:

The equation of the circle of radius 6 and centre at `(3, 5)`

is `(x - 3)^2 + (y- 5) = (6)^2`

Let `S equiv (x - 3)^2 + (y - 5)^2 - 36 = 0`

At point `(-2,-1)`,

`S equiv ( -2-3)^2 + (-1-5)^2 -36`

`=25 + 36 - 36 = 25 > 0`

which represent outside the circle.

At point `(0, 1)`,

`S equiv (0-3)^2+ (1-5)^2 -36`

`= 9 + 16 - 36 = -9 < 0`

which represent inside the circle.

At point `(-1,-2)`,

`S equiv (-1-3)^2+ (-2- 5)^2 -36`

`equiv 16 + 49 - 36 = 29 > 0`

which represent outside the circle.

At point `(2, -1 )`,

`S equiv (2-3)^2 + (-1-5)^2 - 36`

`= 1 +36 -36 =1 > 0`

which represent outside the circle.
Hence, the point `(0, 1)` lies inside the circle `S`.

Correct Answer is `=>` (B) `(0, 1)`

Q 2368678505

Point `(1, 2)` relative to the circle

`x^2 + y^2 + 4x - 2 y - 4 = 0` is a/an
NDA Paper 1 2008

(A)

exterior point

(B)

interior point but not centre

(C)

boundary point

(D)

centre

Solution:

The given equation of circle is

`x^2 + y^2 + 4x - 2 y - 4 = 0`

At `(1,2)`

`(1)^2 + (2)^2 + 4(1)- 2 (2)- 4`

`=1+4+4-4-4=1>0`

Thus, the point `(1, 2)` lies. outside the circle, i.e., this point is an exterior point

Correct Answer is `=>` (A) exterior point

Q 2822623531

Find the position of

A(1,-1)
B(-1,2)

with respect to circle `x^2 + y^2 - 4x + 6y + 4 = 0, `

Solution:

We have `x^2 + y^2 - 4x + 6y + 4 = 0 `⇒ `S = 0,` where `S = x^2 + y^2 - 4x + 6y + 4`

For the point (1, -1), we have `S_A = 1^2 + (-1)^2 - 4.1 + 6. (- 1) + 4 = 1 + 1 - 4 - 6 + 4 = - 4 < 0`

For the point (-1, 2), we have `S_B = (- 1 )^2 + 2^2 - 4. (-1) + 6 .2 + 4 = 1 + 4 + 4 + 12 + 4 = 25 > 0`

Therefore, the point A(1, -1) lies inside the circle whereas B(-1, 2) lies outside the circle.

THE POSITION OF A LINE WITH RESPECT TO A CIRCLE

Let the equation of the circle be

`x^ 2 + y^2 = 32` ........(i)

and the equation of the line be

`y= mx+c`.....(ii)

From (i) and (ii)

`x^2 + (mx +c)^2 = a^2`

`x^2 ( 1 + m^2) + 2 cmx +c^2 - a^2 = 0` ........... (iii)

Case-I :

.When points of intersection are real and distinct, then equation (iii) has two distinct roots.

`:.` Discriminant > 0

or `4m^2c^2 - 4 (1+m^2) (c^2-a^2) > 0`

or `a^2 > c^2/(1+m^2)`

or `a > (|c|)/(sqrt(1+m^2))=` length of perpendicular from `(0, 0)` to `y = mx + c`

`=> a >` length of perpendicular from `(0, 0)` to `y = mx + c`

`=> c^2 < a^2(1+m^2)`

Thus, a line intersects a given circle at two distinct points.

Case-II [Tangent of Circle]

.When the points of intersection are coincident, the equation (iii) has two equal roots

`:. D=0`

`=> a= (|c|)/(sqrt((1+m^2))`

`a =` length of the perpendicular from the point `(0, 0)` to `y = mx + c`

`c^2 = a^2(1+m^2)`

Then line y=mx+c called tangent to the circle

Q 2136678572

If a circle of radius `b` units with centre at `( 0, b)` touches the
line `y = x - sqrt(2)`, then what is the value of `b`?
NDA Paper 1 2016

(A)

`2 + sqrt(2)`

(B)

`2 - sqrt(2)`

(C)

`2sqrt(2)`

(D)

` sqrt(2)`

Solution:

Here, radius of circle = band centre= (0, b)

and the equation ofline touches the circle is

`y = x - sqrt(2)`

:. Perpendicular drawn from centre to the line is

`b =| (0- b - sqrt(2) )/sqrt(2) | = | (-(b + sqrt(2)) )/sqrt(2) |`

` => sqrt(2)b = b + sqrt(2) => b( sqrt(2) - 1) = sqrt(2)`

` => b = sqrt(2)/(sqrt(2) -1) xx (sqrt(2) + 1)/(sqrt(2) + 1 )`

` = (2 + sqrt(2))/(2-1) = (2+ sqrt(2))`

Correct Answer is `=>` (A) `2 + sqrt(2)`

Q 2211523429

A straight line `x = y + 2` touches the circle
`4(x^ 2 + y^2) = r^ 2` . The value of `r` is
NDA Paper 1 2015

(A)

`sqrt(2)`

(B)

`2 sqrt(2)`

(C)

`2`

(D)

`1`

Solution:

Given `x^2 + y^2 = r^2/4`

We know that the line `y = mx + c` meets the circle in

unique real point or touch the circle `x^2 + y^ 2 = r^ 2`, if

` r = | c/sqrt( 1 + m^2)|`

Since, the straight line `x = y + 2` touches the given

circle.

Hence, ` | 2/sqrt(2)| = r/2 => r = 2 sqrt(2)`

Correct Answer is `=>` (B) `2 sqrt(2)`

Q 2358467304

What is the radius of the circle touching `X`-axis at
`(3, 0)` and `Y`-axis at `(0, 3)`?
NDA Paper 1 2011

(A)

`3` units

(B)

`4` units

(C)

`5` units

(D)

`6` units

Solution:

Radius of the circle `= AC = BC`.

`( ·: AC = OB = 3` and 'C = OA = 3`)

`:. ` Radius =` 3` units

Correct Answer is `=>` (A) `3` units

Q 2328167901

If `X`-axis is tangent to the circle `x^2 + y^2 + 2gx + 2fy + k =0`, then which one of
the following is correct?
NDA Paper 1 2009

(A)

`g^2 =k`

(B)

`g^2 =f`

(C)

`f^2 =k`

(D)

`f^2 =g`

Solution:

Since, `X`-axis is a tangent to the given circle, it means
the circle touches the `X`-axis

`:. 2 sqrt(g^2 -k) =0 => g^2 = k`

`text (Alternate Method)`

Equation of circle is `x^2 + y^2 + 2gx + 2fy+ k = 0`

`:.` Radius of circle `= sqrt (g^2 + f^2 - k)`

Equation of `X`-axis is `y = 0`

Centre of circle `( -g, - f)`

Now, length of perpendicular from origin to tangent line i.e., `y = 0`

`=` Radius of circle

`=> (|-f| )/(sqrt ((1)^2 ) ) = sqrt (g^2 + f^2 -K)`

`=> f= sqrt (g^2 + f^2 -K) => f^2=g^2 +f^2 -k`

`=> g^2 = k`

Correct Answer is `=>` (A) `g^2 =k`

Q 2211523429

A straight line `x = y + 2` touches the circle
`4(x^ 2 + y^2) = r^ 2` . The value of `r` is
NDA Paper 1 2015

(A)

`sqrt(2)`

(B)

`2 sqrt(2)`

(C)

`2`

(D)

`1`

Correct Answer is `=>` (B) `2 sqrt(2)`

Q 2136678572

If a circle of radius `b` units with centre at `( 0, b)` touches the
line `y = x - sqrt(2)`, then what is the value of `b`?
NDA Paper 1 2016

(A)

`2 + sqrt(2)`

(B)

`2 - sqrt(2)`

(C)

`2sqrt(2)`

(D)

` sqrt(2)`

Correct Answer is `=>` (A) `2 + sqrt(2)`

Case-III

.When the points of intersection are imaginary. In this case (iii) has imaginary roots

`:. D < 0`

or `a < (|c|)/( sqrt(1+m^2))`

or `a <` length of perpendicular from `(0, 0)` to `y = mx + c`

`=> c^2 > a^2(1+m^2)`

Thus a line does not intersect a circle.

Cutting intercept on axis by a circle

Cirrcle `x^2 + y^2 +2gx + 2fy + c = 0`

if `g^2 > c`, the circle will meet the x-axis at two distinct points.

Length of `x` intercept `= 2sqrt(g^2 - c)`

In the similar manner if `f^2 > c`,

Length of `y` intercept `= 2sqrt (f^2- c)`

`text(Tricks)`

`(i) g^2-c >0 =>` circle cuts the x-axis at two distinct points.

`(ii) g^2 = c =>` circle touches the x-axis.

`(iii) g^2 < c =>` circle lies completely above or below the x-axis i.e. it does not intersect x-axis.

`(iv) f^2 -c >0 =>` circle cuts the y-axis at two distinct points.

`(v) f^2 = c =>` circle touches the y-axis.

`(vi) f^2 < c =>` circle lies completely on the right side or the left side of they-axis i.e. it does not intersect y-axis.

Q 2328167901

If `X`-axis is tangent to the circle `x^2 + y^2 + 2gx + 2fy + k =0`, then which one of
the following is correct?
NDA Paper 1 2009

(A)

`g^2 =k`

(B)

`g^2 =f`

(C)

`f^2 =k`

(D)

`f^2 =g`

Correct Answer is `=>` (A) `g^2 =k`

Q 2281334227

If the centre of the circle passing through the origin is
`(3, 4)`, then the intercepts cut-off by the circle on `X`-axis
and `Y`-axis respectively, are
NDA Paper 1 2015

(A)

`3` units and `4` units

(B)

`6` units and `4` units

(C)

`3` units and `8` units

(D)

`6` units and `8` units

Solution:

We have, centre =` (3, 4)` and radius `= 5`

Equation of circle having centre `(h, k)` and radius `a` is

`(x-h)^2 + (y- k)^2 = a^2`

`=> (x - 3 )^2 + (y - 4)^2 = 25`

For x-intercept

Put `y = 0`, we get `(x- 3 )^2 + 16 = 25`

`=> (x - 3 )^2 = 9`

`=> x-3 = 3` and `-3`

`=> x = 6` and `0`

For y-intercept

Put `x = 0`, we get` 9 + (y - 4)^2 = 25`

`=> y - 4 = 4` and `- 4 => y = 8` and `0`

Hence, the x-intercept is `6` and y-intercept is `8`.

Correct Answer is `=>` (D) `6` units and `8` units

Q 2187178987

Consider a circle passing through the origin and the points `(a, b)`
and `(-b, -a)`.

What is the sum of the squares of the intercepts cut-off
by the circle on the axes ?
NDA Paper 1 2016

(A)

` ( (a^2 + b^2 )/(a^2 - b^2))^2`

(B)

`2 ((a^2 + b^2)/(a - b))^2`

(C)

`4 ((a^2 + b^2)/(a - b))^2`

(D)

None of these

Solution:

Given that circle passes through

`(0, 0), (a, b)` and `(-b,- a)`

Now, equation of circle is

`x^2 + y^2 + 2gx + 2fy + c = 0`

When `x = 0, y = 0 => 0 + 0 + 0 + 0 + c = 0`

`=> c = 0` .......(i)

When `x = a , y = b`

` a^2 + b^2 + 2ga + 2fb = 0 quad [ ∵ c = 0 ]`

` => ga + fb = -1/2 (a^2 + b^2 )` .......(ii)

When `x = - b , y = - a`

` b^2 + a^2 - 2gb - 2fa = 0`

`=> gb + fa = 1/2 (a^2 + b^2)` ... (iii)

On multiplying Eq. (ii) by a and Eq. (iii) by we get

`a^2g + abf = -1/2 (a^2 + b^2 )*a`

`b^2g + abf = 1/2 (a^2 + b^2 )· b`
- - -
_________________

` (a^2 - b^2)g = -1/2 (a^2 + b^2) ( a + b) `

` => g = -1/2 ((a^2 + b^2) (a + b))/((a^2 - b^2) ) `

` => g = -1/2 ((a^2 + b^2)/(a - b))` ............(iv)

Again, on multiplying Eq. (ii) by band Eq. (iii) by a, we
get

` abg + fb^2 = - 1/2 ( a^2 + b^2 ) ·b`

` abg + fa^2 = 1/2 (a^2 + b^2 ) ·a`
- - -
________________

` f(a^2 - b^2) = -1/2 (a^2 - b^2) (b +a)`

` => f = -1/2 (a^2 + b^2)/(b^2 - a^2) ( b + a)`

` f = -1/2 (( a^2 + b^2)/(a - b))` ............(v)

The equation of circle is

`x^2 + y^2 - ((a^2 + b^2)/(a - b) ) x - ( (a^2 + b^2)/(b - a)) y = 0`

Now, for x - intercept, put `y = 0`

`:. x^2 - ((a^2 + b^2)/(a - b)) x = 0 => x = (a^2 + b^2)/(a - b)`

For `y` -intercept, put `x = 0`

`y^2 - ((a^2 + b^2)/(b - a)) y = 0`

`y^2 = (a^2 + b^2)/(b - a)`

Sum of the square of intercepts

`= ( x -` intercept`)^2 + (y -` intercept`)^2`

`= ((a^2 + b^2)/(b - a))^2 + ((a^2 + b^2)/(b - a))^2 = 2 ((a^2 + b^2)/(b - a))^2`

Correct Answer is `=>` (B) `2 ((a^2 + b^2)/(a - b))^2`

Q 2388778607

If the circle
`x^2 + y^2 + 2gx + 2fy + c = 0` (where, `c > 0`)

touches the `Y`-.axis, then which one of the
following is correct?
NDA Paper 1 2008

(A)

`g = - sqrt (c)`

(B)

`g= pm sqrt (c)`

(C)

`f = sqrt (c)`

(D)

`f= pm sqrt (c)`

Solution:

The circle `x^2 + y^2 + 2gx + 2fy + c = 0` touches

`Y`-axis, then `2 sqrt (f^2 - c) = 0`

`=> f^2 = c`

`=> f= pm sqrt (c)`

`text (Alternate Method)`

Equation of Circle is `x^2 + y^2 + 2gx + 2fy+ c = 0`

Here, Radius of Circle`= sqrt (g^2 + f^2 -c )`

Centre of Circle `= (-g , -f)`

Equation of `Y`-axis is `x = 0`.

Now, length of perpendicular from centre of

`Y`-axis i.e., `x = 0` = Radius of circle

`=> (|-g|)/(sqrt((1)^2)) = sqrt (g^2 +f^2 -c)`

`=> g= sqrt (g^2 + f^2 -c)`

`=> g^2 = g^2 + f^2 -c`

`=> f^2 =c`

Correct Answer is `=>` (D) `f= pm sqrt (c)`

Relative position of circles

Q 2136080872

Consider the two circles `(x- 1)^(2) + (y - 3)^( 2) = r^(2)` and

` x^(2) + y ^(2) -8x+ 2y + 8 = 0`

What is the distance between the centres of the two circles?
NDA Paper 1 2016

(A)

`5` units

(B)

`6` units

(C)

`8` units

(D)

`10` units

Solution:

Given circle are

`(x-1)^(2) + (y-3)^(2) =r^(2) ... (i)`

and `x^( 2) + y^( 2) - 8x + 2y + 8 = 0 ... (ii)`

Centre of circle (i) `= (1, 3)` and radius `r_(1) = r`

Centre of circle (ii) `= (4, -1)`

and radius `(r_(2) ) = sqrt(16 + 1 - 8) = sqrt(9) = 3`

Distance between centres `=sqrt ((1- 4)^(2) + (3+ 1)^(2))`

`= sqrt(9 + 16) = sqrt(25) = 5` units

Correct Answer is `=>` (A) `5` units

Q 2136080872

Consider the two circles `(x- 1)^(2) + (y - 3)^( 2) = r^(2)` and

` x^(2) + y ^(2) -8x+ 2y + 8 = 0`

What is the distance between the centres of the two circles?
NDA Paper 1 2016

(A)

`5` units

(B)

`6` units

(C)

`8` units

(D)

`10` units

Correct Answer is `=>` (A) `5` units

Circles do not intersect

`C_1C_2>r_1+r_2`
Four common tangent can be drawn- two direct `& ` two transverse

Circles touch each other externally

`C_1C_2=r_1+r_2`

Three common tangents can be drawn .

Circles intersect in two points

`|r_1- r_2| < C_1C_2 < r_1+r_2`

Two common tangents can be drawn

Q 2116180970

Consider the two circles `(x- 1)^(2) + (y - 3)^( 2) = r^(2)` and

` x^(2) + y ^(2) -8x+ 2y + 8 = 0`

If the circles intersect at two distinct points, then which
one of the following is correct?
NDA Paper 1 2016

(A)

`r = 1`

(B)

`1 < r < 2`

(C)

`r = 2`

(D)

`2 < r < 8`

Solution:

Given circle are

`(x-1)^(2) + (y-3)^(2) =r^(2) ... (i)`

and `x^( 2) + y^( 2) - 8x + 2y + 8 = 0 ... (ii)`

Centre of circle (i) `= (1, 3)` and radius `r_(1) = r`

Centre of circle (ii) `= (4, -1)`

and radius `(r_(2) ) = sqrt(16 + 1-8) = sqrt(9) = 3`

Distance between centres `=sqrt ((1- 4)^(2) + (3+ 1)^(2))`

`= sqrt(9 + 16) = sqrt(25) = 5` units

Condition for two circles intersect at two distinct
points,

Distance between centre < Sum of radius

` 5 < r + 3 => 2 < r`

But r > 1 because `5 < 1 + 3`

Correct Answer is `=>` (B) `1 < r < 2`

Circles touch each other internally

`C_1C_2=|r_1-r_2|`

Only one common tangent can be drawn

One circle lie completely inside other

`0<= C_1C_2 < |r_1-r_2|`

No common tangent can be drawn.

miscellaneous

Q 2713691549

What is the maximum area of a triangle that can be inscribed in a circle of radius `a` ?
NDA Paper 1 2017

(A)

`(3a^2)/4`

(B)

`a^2/2`

(C)

`(3 sqrt 3 a^2)/4`

(D)

`(sqrt 3 a^2)/4`

Solution:

the maximum area of a triangle that can be inscribed in a circle of radius `a`, triangle should be equilateral.

So area of equilateral triangle `A= (sqrt 3)/4 a^2`

Correct Answer is `=>` (D) `(sqrt 3 a^2)/4`

Q 2338567402

What is the equation to circle which touches both
the axes and has centre on the line `x + y = 4`?
NDA Paper 1 2010

(A)

`x^2 + y^2 - 4x + 4y + 4 = 0`

(B)

`x^2 + y^2 -4x- 4y + 4 = 0`

(C)

`x^2 + y^2 + 4x- 4y- 4 = 0`

(D)

`x^2 + y^2 + 4x + 4y- 4 = 0`

Solution:

We know that, the equation of circle, which touches
both the axes, is

`x^2 + y^2 - 2rx- 2ry + r^2 = 0` ... (i)

The centre `(r, r)` of this circle lies on the line `x + y = 4`.

`:. r+r=4 => r=2`

On putting the value of `r` in Eq. (i), we get

`x^2 + y^2 - 4x - 4y + 4 = 0`

which is the required equation of circle.

Correct Answer is `=>` (B) `x^2 + y^2 -4x- 4y + 4 = 0`