Question

The equation of a circle in parametric form is given by $x=a\cos \theta ,y=a\sin \theta$. If $a=2$, the equation of the chord AB of the above circle, whose mid-point is at a perpendicular distance of $3$ from the point $P(h,k)$ on the circle is:

• A. $x+y+h+k=4$
• B. $hx-ky=4$
• C. $hx+ky=4$
• D. $hx+ky=-2$

Answer 1

Answer: (D)

$hx+ky=-2$

The equation of the circle is $x^2+y^2=4$

From what is given it is clear that the One joining $(h,k)$ and the mid-point of the chord AB passes through the origin.

$\therefore$ if the mid-point is $(x_1,y_1),MO=1,OP=2$

$\Rightarrow \frac{1-h+2.x_1}{3}=0,\frac{1.k+2.y_1}{3}=0$

$\Rightarrow M$ is $(-\frac{h}{2},-\frac{k}{2})$

Slope of AB is $-\frac{h}{k}(\because itis\perp OP)$

$\therefore$ equation of AB is $y+\frac{k}{2}=-\frac{h}{k}(x+\frac{h}{2})$

i.e., $hx+ky=-\frac{1}{2}(h^2+k^2)=-\frac{4}{2}$ (as P lies on the circle) $=-2$