Question

If the equation of tangent to a circle at point (3, 5) is $2x-y-1=0$ and its centre lies on $x+y=5,$ then the equation of circle is

• A. $x^2+y^2+6x-16y+28=0$
• B. $x^2+y^2-6x-16y+28=0$
• C. $x^2+y^2+6x+6y+28=0$
• D. $x^2+y^2-6x-6y-28=0$

Answer 1

Answer: (A)

Equation of line which is perpendicular to the tangent
$2x-y-1=0$ is $x+2y+k=0$ .
Equation of tangent passes through the point (3,5).
$\therefore$ $3+2\times 5+k=0$
$\Rightarrow$ $k=-13$
$\therefore$ Centre of circle will lie on the perpendicular line
$x+2y=13$ .
Also, centre of circle lies on the line $x+y=5$ .
So, on solving these two equations, we get
$x=-3,y=8$
Centre $=(-3,8)$
and radius $=\sqrt{{(3+3)}^2+{(5-8)}^2}$
$=\sqrt{6^2+3^2}$
$=\sqrt{45}$
$\therefore$ Equation of circle is ${(x+3)}^2+{(y-8)}^2={(\sqrt{45})}^2$
$\Rightarrow$ $x^2+9+6x+y^2+64-16y=45$
$\Rightarrow$ $x^2+y^2+6x-16y+28=0$