Question

The equation of the circle which has a tangent $2x-y-1=0$ at $(3,5)$ on it and with the center on $x+y=5$, is

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

Answer 1

Answer: (A)

$x^2+y^2+6x-16y+28=0$

Clearly, the center of the circle lies on the line through the point $(3,5)$ and $\perp$ to the tangent $2x-y-1=0$

the equation of such line is

$(y-5)=-\frac{1}{2}(x-3)\Rightarrow x+2y=13$ ...(1)

Also, it is given that center lies on the line $x+y=5$ ...(2)

Solving (1) and (2), we obtain the coordinate of the center of the circle as $C\equiv (-3,8)$

Alos, radius of the circle $=\sqrt{36+9}=\sqrt{45}$

$\therefore$ Equation of the circle is

${(x+3)}^2+{(y-8)}^2={\left(\sqrt{45}\right)}^2\Rightarrow x^2+y^2+6x-16y+28=0$