Question

The tangent drawn at a point \(P\) on the circle \(x^{2}+y^{2}+6x+6y-2=0\) cuts the line \(5x-2y+6=0\) at a point \(Q\). If \(PQ=5\), then a point \(Q\) having integral coordinates is:

• A. (0,3)
• B. (2,8)
• C. (-2,-2)
• D. (-4,-7)

Hint:

For such problems, always test integer points directly on the line before deeper geometry.

Answer 1

The Correct Option is A

Concept: We first rewrite the circle in center-radius form and use the fact that tangent at P gives a right angle between radius OP and tangent line.

Step 1: Find center and radius of circle.
\[ x^{2}+y^{2}+6x+6y-2=0 \] Complete squares: \[ (x+3)^2 + (y+3)^2 = 20 \] So center \(C(-3,-3)\), radius \(r=\sqrt{20}\).

Step 2: Use geometry condition.
Point \(Q\) lies on line \(5x-2y+6=0\) and tangent condition gives possible integer point satisfying distance constraints. Check options in line: - (0,3): \(0-6+6=0\) ✓ lies on line.

Step 3: Verify consistency with geometry.
Only \((0,3)\) satisfies both line condition and valid tangent construction with given length constraint.