Question

The equation of the circle whose center is $(3,-1)$ and which cuts off an intercept of length $6$ from the line $2x-5y+18=0$, is

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

Answer 1

Answer: (D)

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

Let $C$ be the center of the circle, then $C\equiv (3,-1)$

Equation of line $AB$ is $2x-5y+18=0$

and $AB=6$

$\therefore AL=3$

$CL=$ length of the $\perp$ from $C$ on $AB$

$=\frac{|2\times 3-5(-1)+18|}{\sqrt{{\left(2\right)}^2+{(-5)}^2}}=\sqrt{29}$

$\therefore$ radius of the circle $AC=\sqrt{{AL}^2+{CL}^2}=\sqrt{3^2+29}=\sqrt{38}$

Thus, the equation of the required circle is

${(x-3)}^2+{(y+1)}^2=38\Rightarrow x^2+y^2-6x+2y-28=0$