Question

The equation of a circle whose centre is $(3,-1)$ and which intercept chord of $6$ units length on straight 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. Both of them
• D. none of these

Answer 1

The correct option is A $x^2+y^2-6x+2y-28=0$

centre $=(3,-1)=(x,y)$

straight line $=2x-5y+18=0$ (Chord of circle)\

perpendicular distance from centre to the straight line

$\Rightarrow \frac{2x_1-5y_1+18}{\sqrt{2^2+5^2}}$

$\Rightarrow \frac{2\left(3\right)-5(-1)+18}{\sqrt{29}}$

$\Rightarrow \frac{29}{\sqrt{29}}=\sqrt{29}$

$\therefore$ perpendicular distance from centre to chord is

chord length $=6$ units (given)

${\left(\frac{\text{chord length}}{2}\right)}^2+\text{(perpendicular distance to chord}{)}^2=(\text{radius}{)}^2$

${\left(\frac{6}{2}\right)}^2+{\left(\sqrt{29}\right)}^2=(\text{radius}{)}^2$

$\Rightarrow 9+29=(\text{radius}{)}^2$

$(\text{radius}{)}^2=38$

$\therefore$ equation of centre is $(x-3{)}^2+(y+1{)}^2=(\text{radius}{)}^2$

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

$\Rightarrow x^2-6x+y^2+2y+10=38$

$\Rightarrow x^2+y^2-6x+2y-2=0$