<p>If the area of the quadrilateral formed by the tangents from the origin to the circle \(x^2 + y^2 + 6x - 10y + c = 0\) and the radii corresponding to the points of contact is 15, then a value of <i>c</i> is:</p>
Step-by-Step Solution
Key Concept: When tangents are drawn from an external point to a circle, they form a quadrilateral with the two radii to points of contact. This quadrilateral has a specific area formula in terms of the tangent length and radius.
<p><strong>Step 1:</strong> Rewrite the circle equation in standard form.</p><p>$x^2 + y^2 + 6x - 10y + c = 0$</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: $O' = (-3, 5)$, Radius: $r = \sqrt{34 - c}$</p><p><strong>Step 2:</strong> Find the tangent length from origin O(0,0) to the circle.</p><p>Distance from O to O': $d = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$</p><p>If $PT$ is the tangent from O to circle at point $T$, then:</p><p>$OT^2 = d^2 - r^2 = 34 - (34 - c) = c$</p><p>So tangent length: $PT = \sqrt{c}$</p><p><strong>Step 3:</strong> Identify the quadrilateral geometry.</p><p>The quadrilateral formed by the two tangents and two radii to the points of contact is a kite. Let the two tangent points be $P$ and $Q$. The quadrilateral $OPTQ$ has:</p><p>- Right angles at $P$ and $Q$ (radius ⊥ tangent)</p><p>- $OP = OQ = \sqrt{c}$ (equal tangent lengths)</p><p>- $O'P = O'Q = r = \sqrt{34-c}$ (radii)</p><p><strong>Step 4:</strong> Calculate the area of quadrilateral.</p><p>The quadrilateral consists of two congruent right triangles: $\triangle OPO'$ and $\triangle OQO'$</p><p>Area of $\triangle OPO' = \frac{1}{2} \times OP \times O'P = \frac{1}{2} \times \sqrt{c} \times \sqrt{34-c}$</p><p>Total area of quadrilateral:</p><p>$A = 2 \times \frac{1}{2} \times \sqrt{c} \times \sqrt{34-c} = \sqrt{c(34-c)} = \sqrt{34c - c^2}$</p><p><strong>Step 5:</strong> Use the given condition that area = 15.</p><p>$\sqrt{34c - c^2} = 15$</p><p>$34c - c^2 = 225$</p><p>$c^2 - 34c + 225 = 0$</p><p><strong>Step 6:</strong> Solve the quadratic equation.</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:</strong> Verify validity.</p><p>For the circle to exist: $34 - c > 0$, so $c < 34$</p><p>Both $c = 9$ and $c = 25$ satisfy this condition.</p><p>However, $c = 9$ is listed as option (a).</p><p>∴ Answer: a</p>
Correct Answer: a