Question

Obtain the equation of the circle orthogonal to both the circles $x^2+y^2+3x-5y+6=0$ and $4x^2+4y^2-28x+29=0$ and whose centre lies on the line $3x+4y+1=0$

Answer 1

Given circles are $x^2+y^2+3x-5y+6=0$.................(i)
and $4x^2+4y^2-28x+29=0$
or $x^2+y^2-7x+\frac{29}{4}=0$..............(ii)
Let the required circle be $x^2+y^2+2gx+2fy+c=0$....................(iii)
Since circle (iii) cuts circles (i) and (ii) orthogonally
$\therefore 2g\left(\frac{3}{2}\right)+2f(-\frac{5}{2})=c+6$ or $3g-5f=c+6$ ..............(iv)
and $2g(-\frac{7}{2})+2f.0=c+\frac{29}{4}or-7g=c+\frac{29}{4}.............\left(v\right)$
From (iv) & (v) we get $10g-5f=-\frac{5}{4}$ or 40g - 20f = -5 ..............(vi)
Given line is $3x+4y=-1$ .............(vii)
Since centre (-g, -f) of circle (iii) lies on line (vii) .............(viii)
Solving (vi) & (viii) we get $g=0,f=\frac{1}{4}$ $\therefore from\left(5\right),c=-\frac{29}{4}$
$\therefore$ From (iii) required circle is $x^2+y^2+\frac{1}{2}y-\frac{29}{4}=0$ or $4(x^2+y^2)+2y-29=0$