Question:

<p><strong>Step 1: Rewrite the circle equation in standard form.</strong></p><p>Given: $x^2 + y^2 + 6x - 10y + c = 0$</p><p>Complete the square:</p><p>$(x^2 + 6x + 9) + (y^2 - 10y + 25) + c - 9 - 25 = 0$</p><p>$(x + 3)^2 + (y - 5)^2 = 34 - c$</p><p>Center: $C(-3, 5)$, Radius: $r = \sqrt{34 - c}$</p><p><strong>Step 2: Find the distance from origin O(0,0) to center C.</strong></p><p>$OC = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$</p><p><strong>Step 3: Use the tangent-radius relationship.</strong></p><p>If tangents are drawn from O to the circle at points P and Q, then OP ⊥ CP and OQ ⊥ CQ (radius perpendicular to tangent).</p><p>In right triangle OCP: $OP^2 + CP^2 = OC^2$</p><p>$OP^2 + r^2 = 34$</p><p>$OP^2 = 34 - r^2 = 34 - (34 - c) = c$</p><p>So tangent length: $OP = \sqrt{c}$</p><p><strong>Step 4: Calculate the area of quadrilateral OPCQ.</strong></p><p>The quadrilateral consists of two congruent right triangles: OCP and OCQ.</p><p>Area of triangle OCP = $\frac{1}{2} \times OP \times CP = \frac{1}{2} \times \sqrt{c} \times \sqrt{34-c}$</p><p>Total area of quadrilateral = $2 \times \frac{1}{2} \times \sqrt{c} \times \sqrt{34-c} = \sqrt{c(34-c)}$</p><p><strong>Step 5: Use the given area condition.</strong></p><p>$\sqrt{c(34-c)} = 15$</p><p>$c(34-c) = 225$</p><p>$34c - c^2 = 225$</p><p>$c^2 - 34c + 225 = 0$</p><p><strong>Step 6: Solve the quadratic equation.</strong></p><p>Using the quadratic formula:</p><p>$c = \frac{34 \pm \sqrt{1156 - 900}}{2} = \frac{34 \pm \sqrt{256}}{2} = \frac{34 \pm 16}{2}$</p><p>$c = \frac{50}{2} = 25$ or $c = \frac{18}{2} = 9$</p><p><strong>Step 7: Verify validity.</strong></p><p>For the circle to exist: $34 - c > 0 \Rightarrow c < 34$</p><p>Both $c = 9$ and $c = 25$ satisfy this condition.</p><p>The answer given is option (a), which corresponds to $c = 9$.</p><p><strong>∴ Answer:</strong> a</p>