Question

The line $2x-y+1=0$ is a tangent to the circle at the point $(2,5)$ and the centre of the circle is on the line $x+y=9$. The equation of the circle is

• A. $x^2+y^2-12x+6y+5=0$
• B. $x^2+y^2-12x-6y-5=0$
• C. $x^2+y^2+12x+6y+15=0$
• D. $x^2+y^2-12x-6y+25=0$

Answer 1

Answer: (D)

$x^2+y^2-12x-6y+25=0$

Clearly, the centre of the circle is the point of intersection of the line $x+y=9$ and the normal at $P$ i.e. a line passing through $P$ and perpendicular to the tangent at $P$.

Slope of the line $2x-y+1$ is $2$

$\therefore$ Slope of the line $CP$ is $-\frac{1}{2}$
The equation of $CP$ is
$y-5=-\frac{1}{2}(x-2)\Rightarrow x+2y=12...\left(i\right)$
Solving this equation with $x+y=9$, we obtain that the coordinates of $C$ are $(6,3).$
Now,
Radius of the required circle = $CP=\sqrt{(6-2{)}^2+(5-3{)}^2}=\sqrt{20}$
Hence, the equation of the required circle is
$(x-6{)}^2+(y-3{)}^2=20$ or, $x^2+y^2-12x-6y+25=0$