Question

Find the equation of the circle orthogonal to the circles $x^2+y^2+3x-5y+6=0and4x^2+4y2-28x+29=0$ and whose center lies on the line 3x + 4y + 1 = 0.

• A. $x^2+y^2+\frac{y}{4}-\frac{29}{4}=0$
• B. $2x^2+2y^2+y-29=0$
• C. $8x^2+8y^2+2y-29=0$
• D. $4x^2+4y^2+2y-29=0$

Answer 1

The correct option is D

$4x^2+4y^2+2y-29=0$

Let the required circle be

$x^2+y^2+2gx+2fy+c=0$ .............(1)

Given circles are

$x^2+y^2-3x-5y+6=0$ ............. (2)

$4x^2+4y^2-28x+29=0$

$x^2+y^2-7x+\frac{29}{4}=0$ .............(3)

Since circles (1)$\&$(2) and (1)$\&$(3) are orthogonal to each other then

$2g_1{g}_2+2f_1{f}_2=c_1+c_2$

$2(-g)\left(\frac{-3}{2}\right)+2(-f)\left(\frac{+5}{2}\right)=c+6$

$3g-5f=c+6$ .............(4)

and $2(-g)\left(\frac{7}{2}\right)+2(-f)\left(0\right)=c+\frac{29}{4}$

$-7g=c+\frac{29}{4}$ ...........(5)

from equation (4) and (5)

$3g-5f=-7g-\frac{29}{4}+6$

$10g-5f=\frac{-5}{4}$

$2g-f=\frac{1}{4}$ .............(6)

centre (-g,-f) lies on the line $3x+4y+1=0$

$-3g-4f=-1$ .............(7)

Solving equation (6) and (7)

We get, $g=0andf\frac{1}{4}$

From equation (5) $c=\frac{29}{4}$

Substituting the values of g, f and c in equation (1)

Required equation of circle is

$x^2+y^2+2\left(0\right)x+2\left(\frac{1}{4}\right)y+\left(\frac{-29}{4}\right)=0$

$4x^2+4y^2+2y-29=0$