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 \(Q\).
Since \(Q\) lies on the \(y\)-axis,
\[ x=0 \]
Substitute \(x=0\) in
\[ 5x-2y+6=0 \]
\[ -2y+6=0 \]
\[ y=3 \]
So,
\[ Q=(0,3) \]

Step 2: Use tangent length formula.
For the circle
\[ x^2+y^2+6x+6y-2=0 \]
The length of tangent from an external point \(Q(x_1,y_1)\) is
\[ \sqrt{S_1} \]
where
\[ S_1=x_1^2+y_1^2+6x_1+6y_1-2 \]

Step 3: Substitute \(Q=(0,3)\).
\[ S_1=0^2+3^2+6(0)+6(3)-2 \]
\[ =9+18-2 \]
\[ =25 \]
Therefore,
\[ PQ=\sqrt{25}=5 \]

Step 4: Final conclusion.
Hence,
\[ \boxed{5} \]