Question

If the tangent at the point $P$ on the circle $x^2+y^2+6x+6y-2=0$ meets the straight line $5x-2y+6=0$ at point $Q$ on the $y$-axis, then the length of $PQ$ is

• A. $2$
• B. $5$
• C. $2\sqrt{5}$
• D. $3\sqrt{5}$

Answer 1

The correct option is B $5$

Equation of the circle is $x^2+y^2+6x+6y-2=0$ ...(i)

and equation of the straight line is $5x-2y+6=0$$\Rightarrow y=\frac{5}{2}x+3$ ...(ii)

We know that the standard equation of the circle is $x^2+y^2+2gx+2fy+c=0$

Comparing Eq.(i) with the standard equation, we get $g=3,f=3$ and $c=-2.$

We also know that the coordinates of the centre of the circle $O(-g.-f)=(-3,-3)$

and its radius $\left(r\right)=\sqrt{g^2+f^2-c}=\sqrt{9+9+2}=\sqrt{20}$

We also know that the standard equation of a straight line is $y=mx+c.$

Comparing Eq. (ii) with the standard equation, we get $m=\frac{5}{2}$ and $c=3.$

Thus, the straight line intersects the $y$-axis at $Q(0,3).$

We also, know that the distance between the point $Q$ and centre $O$ is $\left(QO\right)=\sqrt{{(0+3)}^2+{(3+3)}^2}=\sqrt{45}$

Therefore, the distance between points $P$ and $Q$ is $\left(PQ\right)=\sqrt{{\left(QO\right)}^2-{\left(r\right)}^2}=\sqrt{45-20}=\sqrt{25}=5$