Question

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

• A. \(5\)
• B. \(6\)
• C. \(4\)
• D. \(3\)

Hint:

Length of tangent from point \((x_1,y_1)\) to the circle \(S=0\) is \(\sqrt{S_1}\), where \(S_1\) is obtained by substituting \((x_1,y_1)\) in the circle equation.

Answer 1

The Correct Option is A

Step 1: Find point $Q$ on the $y$-axis.
On the $y$-axis $x=0$. Substituting into $5x-2y+6=0$ gives $-2y+6=0$, so $y=3$. Thus $Q=(0,3)$.
Step 2: Recognise $PQ$ as a tangent length.
$P$ is the point of tangency and $PQ$ is the tangent segment from the external point $Q$ to the circle.
Step 3: Recall the tangent-length formula.
For the circle $x^2+y^2+6x+6y-2=0$, the squared tangent length from $(x_1,y_1)$ is $S_1=x_1^2+y_1^2+6x_1+6y_1-2$.
Step 4: Substitute $Q=(0,3)$.
$S_1=0+9+0+18-2=25$.
Step 5: Take the square root.
$PQ=\sqrt{S_1}=\sqrt{25}=5$.
Step 6: State the length.
So $PQ=5$.
\[ \boxed{5} \]