Mathematics previous year question of Circle for NDA

Previous Year NDA Question of Circle :

Previous Year NDA Question of 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 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
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 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`
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 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 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 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 1619112010

Consider the parametric equation `x = ( a(1 - t^2) )/( 1 +t^2) , y = (2at)/( 1 +t^2)`

What does the equation represent?
NDA Paper 1 2015
(A)

It represents a circle of diameter a

(B)

It represents a circle of radius a

(C)

It represents a parabola

(D)

None of the above

Solution:

We have, `x = ( a(1 - t^2) )/( 1 +t^2) ,` and ` y = (2at)/( 1 +t^2)`

` x^2 + y^2 = ( a^2 ( 1 - t^2)^2)/( 1 +t^2)^2 + (4a^2t^2)/( 1 +t^2)^2`

` = a^2/( 1 +t^2)^2 [ ( 1-t^2)^2 =4t^2]`

` = a^2/( 1 +t^2)^2 (1 + t^4 - 2t^2 + 4t^2) = a^2/( 1 +t^2)^2 ( 1 +t^2)^2`

` => x^2 + y^2 = a^2` ...........(1)

Which is an equation of circle with radius a.
Correct Answer is `=>` (B) It represents a circle of radius a
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 1639178012

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

The two circles touch each other, if
NDA Paper 1 2014
(A)

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

(B)

` 1/c = 1/a^2 + 1/b^2 `

(C)

` c = 1/a^2 + 1/b^2 `

(D)

` c = 1/ (a^2 - b^2)`

Solution:

Two circles touch each other, iff

Distance between two centres = Sum of radius of two

circles

` => sqrt( a^2 + b^2) = sqrt( a^2 + c ) + sqrt( b^2 + c )`

On squaring both sides, we get

`a^2 + b^2 = a^2 - c + b^2 - c + 2 sqrt(( a^2 + c )( b^2 + c ))`

` => c = sqrt(( a^2 + c )( b^2 + c ))`

Again, squaring both sides, we get

`c^ 2 = a^2b^2 - a^2c - b^ 2c + c^ 2`

`=> a^2 b^2 =(a^2 + b^ 2 ) c => 1/c = 1/a^2 + 1/b^2 `
Correct Answer is `=>` (B) ` 1/c = 1/a^2 + 1/b^2 `
Q 1713856749

Consider the spheres

`x^ 2 + y^ 2 + z^2 - 4 y + 3= 0` and
`x^ 2 + y^ 2 + z^ 2 + 2x + 4z - 4 = 0`.

Consider the following statements

I. The two spheres intersect each other.
II. The radius of first sphere is less than that of second
sphere.

Which of the above statements is/are correct?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Given equations of sphere,

`x^2 + y^2 + z^2 -4y + 3 = 0` ....(i)

and `x^ 2 + y^2 + z^2 + 2x + 4z - 4 = 0` .... (ii)

Compair with the standard equation of sphere,

`x^2 + y^2 + z^2 + 2 ux + 2vy + 2wz + d = 0`, we get

`u_1 = 0, v_1 =-2, w_1 = 0` and `d_1 = 3` (for first sphere)

and `u_2 = 1, v_2 = 0, w_2 = 2` and `d_ 2 =- 4` (for second sphere)

I. Radius of 1st sphere,

` r_1 = sqrt( u_1^2 + v_1^2 + w_1^2 - d_1)`

` = sqrt( (0)^2 + (2)^2 + (0)^2 - (3))`

` = sqrt(4 - 3) = sqrt(1) = 1`

Fladius of 2nd sphere,

` r_2 = sqrt( u_2^2 + v_2^2 + w_2^2 - d_2)`

` = sqrt ((-1)^2 + (0)^2 + (-2)^2 + (4))`

` = sqrt(1 + 4 + 4) = sqrt(9) = 3`

Now, `r_1 + r_2 = 1 + 3 = 4`

` c_1 c_1 < r_1 + r_2`

So, both sphere intersect each other.

2. We have, radius of 1st sphere `(r_1) = 1`

and radius of 2nd sphere `(r_2 ) = 3`

i.e.,` r_1 < r_2`

So, the radius of first sphere is less than that of second

sphere.
Correct Answer is `=>` (C) Both 1 and 2
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 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 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 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 2337180982

What is the set of points `(x, y)` satisfying the
equations `x^2 + y^2 = 4` and `x + y = 2`?
NDA Paper 1 2011
(A)

`{(2, 0), (-2, 0), (0, 2)}`

(B)

`{(0, 2), (0, -2)}`

(C)

`{(0, 2), (2, 0)}`

(D)

` { (2, 0), (-2, 0), (0, 2), (0, -2)}`

Solution:

The given equations are

`x^2 + y^2 =4` ... (i)

and `x + y = 2` ... (ii)

These equations are satisfied by `(2, 0)` and `(0, 2)`.

Hence, the required set is `{(0, 2), (2, 0)}`.
Correct Answer is `=>` (C) `{(0, 2), (2, 0)}`
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`
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
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 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 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 2318478300

`ABC` is an equilateral triangle inscribed in a circle
of centre `O` and radius `5 cm`. Let the diameter
through `C` meet the circle again at `D`.
NDA Paper 1 2008

Assertion : ` AD * BD < OB 8 OC`

Reason : `2 (AD^2 + BD^2 ) = CD^2 = 100 sq cm`

(A) Both A and R individually true and R is the correct explanation of A
(B) Both A and R are individually true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Solution:

From figure, `Delta ABC` is an equilateral triangle.

Then, `Delta ODA` is also an equilateral triangle.

Similarly, `Delta OBD` is also an equilateral triangle.

`:. OB = OA = AD = BD = 5`

`AD*BD=25`

and `OB*OC = 25`

`=> AD*BD =OB*OC`

and `2 (AD^2 + BD^2 ) = 2 (25 + 25)`

`= 100 = CD^2`
Correct Answer is `=>` (D)
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 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 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)`
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`

 
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