Question

A tangent drawn through the point $(2,-1)$ to a circle meets it at $(2,3)$. If radius of the circle is $3$ units, then equation of the circle can be

• A. $x^2+y^2-10x-6y+25=0$
• B. $x^2+y^2+2x-6y+1=0$
• C. $x^2+y^2+2x+6y+1=0$
• D. $x^2+y^2-6x-10y+19=0$

Answer 1

Answer: (A), (B)

The correct options are
A $x^2+y^2-10x-6y+25=0$
B $x^2+y^2+2x-6y+1=0$

The line passing through $(2,3)$ and $(2,-1)$ is
$x=2$
Equation of the normal at $(2,3)$ is
$y=\lambda$
This passes through $(2,3)$, so the normal is
$y=3$


Let the centre of the circle be $C=(h,3)$
Now, distance from centre to tangent is equal to radius
$\left|\frac{h-2}{\sqrt{1^2}}\right|=3\Rightarrow |h-2|=3\Rightarrow h-2=\pm 3\Rightarrow h=2\pm 3\Rightarrow h=-1,5$

Therefore, the required equations of circle are
$(x-5{)}^5+(y-3{)}^2=9$ and
$(x+1{)}^2+(y-3{)}^2=9$
$\Rightarrow x^2+y^2-10x-6y+25=0$ and
$x^2+y^2+2x-6y+1=0$