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))`

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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`

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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

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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`

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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`

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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`

---

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)`

---

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`.

---

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.

---

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)`

---

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 `

---

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.

---

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`

---

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`.

---

Solution:

Radius of the circle `= AC = BC`.

`( ·: AC = OB = 3` and 'C = OA = 3`)

`:. ` Radius =` 3` units

---

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`

---

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)}`.

---

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.

---

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.

---

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.

---

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`

---

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`

---

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`

---

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`

---

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

---

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`

---

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`

---