Question

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

• A. $4$
• B. $2\sqrt{5}$
• C. $5$
• D. $3\sqrt{5}$

Answer 1

The correct option is C

$5$

Explanation for the correct answer.

Step 1: Find the coordinates of $Q$.

As $Q$ is a point on $y-axis$ so, the coordinates will be $\left(0,y\right)$ and it will satisfy the equation $5x-2y+6=0$.

So,

$\begin{array}{rcl}5\left(0\right)-2y& =& 6\\ \Rightarrow y& =& 3\end{array}$

So, the coordinates of $Q$ will be $\left(0,3\right)$.

Step 2: Find the radius and Centre of the circle.

The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$, where the radius is $\sqrt{\left(g^2+f^2-c\right)}$ and the Centre is $\left(-g,-f\right)$.

Comparing the given equation of circle with the general equation we get, $g=3,f=3,c=-2$

So, the Centre will be $\left(-3,-3\right)$ and the radius will be $\sqrt{\left(9+9+2\right)}=2\sqrt{5}$

Step 3: Find the length of $PQ$.

Distance between the point $Q\left(0,3\right)$ and Centre $\left(-3,-3\right)$ will be,

$\begin{array}{rcl}OQ& =& \sqrt{{\left(3+0\right)}^2+{\left(-3-3\right)}^2}\left[byd=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right]\\ & =& \sqrt{9+36}\\ & =& 3\sqrt{5}\end{array}$

$OP$ is the radius, so it is perpendicular to $PQ$.

By applying Pythagoras theorem in $∆OPQ$, we get

$\begin{array}{rcl}PQ& =& \sqrt{O{Q}^2-O{P}^2}\\ & =& \sqrt{45-20}\\ & =& 5\end{array}$

Hence, option (C) is correct.