Question

The equation of circle which passes through focus of parabola $x^2=4y$ and touches it at $(6,9)$ is

• A. $x^2+y^2+18x-28y+27=0$
• B. $x^2+y^2+24x-27y+26=0$
• C. $x^2+y^2+48x-12y+11=0$
• D. $x^2+y^2+18x-22y+21=0$

Answer 1

The correct option is A $x^2+y^2+18x-28y+27=0$


$\frac{dy}{dx}{|}_{(6,9)}=\frac{2x}{4}=3$
Equation of tangent : $(y-9)=3(x-6)$
$\Rightarrow 3x-y=9$
Taking $(6,9)$ as point on circle and line as equation of tangent, using the concept of family of circles
$S+\lambda L=0$
$\Rightarrow (x-6{)}^2+(y-9{)}^2+\lambda (3x-y-9)=0\cdots \left(1\right)$
Required circle passes through $(0,1).$
$\therefore 36+64+\lambda (-10)=0$
$\Rightarrow \lambda =10$
Putting $\lambda =10$ in $\left(1\right),$ we get
$(x-6{)}^2+(y-9{)}^2+10(3x-y-9)=0$
$\Rightarrow x^2+y^2+18x-28y+27=0$