Question

Let the tangents at the points A(4, –11) and B(8, –5) on the circle \(x^2+y^2-3x+10y-15=0\), intersect at the point C. Then the radius of the circle, whose centre is C and the line joining A and B is its tangent, is equal to

• A. √13
• B. \( \frac{3 \sqrt{3}}{3}\)
• C. \( \frac{2 \sqrt{13}}{3}\)
• D. 2√13

Answer 1

The Correct Option is C

We are given the equation of the circle: \[ x^2 + y^2 - 3x + 10y - 15 = 0. \] First, rewrite the equation in standard form by completing the square.

Step 1: Completing the square. For \( x \), we have: \[ x^2 - 3x = (x - \frac{3}{2})^2 - \frac{9}{4}. \] For \( y \), we have: \[ y^2 + 10y = (y + 5)^2 - 25. \] Substitute these into the original equation: \[ (x - \frac{3}{2})^2 - \frac{9}{4} + (y + 5)^2 - 25 - 15 = 0. \] Simplifying: \[ (x - \frac{3}{2})^2 + (y + 5)^2 = \frac{9}{4} + 40 = \frac{169}{4}. \] Thus, the equation of the circle is: \[ (x - \frac{3}{2})^2 + (y + 5)^2 = \left( \frac{\sqrt{169}}{2} \right)^2, \] which shows that the center of the circle is \( \left( \frac{3}{2}, -5 \right) \) and the radius is \( \frac{13}{2} \). 

Step 2: Using the formula for the length of the tangent from an external point. The length of the tangent from a point \( (x_1, y_1) \) to a circle with center \( (h, k) \) and radius \( r \) is given by: \[ l = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2}. \] For point \( A(4, -11) \), the tangent length is: \[ l_A = \sqrt{(4 - \frac{3}{2})^2 + (-11 + 5)^2 - \left( \frac{13}{2} \right)^2}. \] Simplifying: \[ l_A = \sqrt{\left( \frac{5}{2} \right)^2 + (-6)^2 - \left( \frac{13}{2} \right)^2} = \sqrt{\frac{25}{4} + 36 - \frac{169}{4}} = \sqrt{\frac{25 + 144 - 169}{4}} = \sqrt{\frac{0}{4}} = 0. \] Thus, the length of the tangent is \( \frac{2 \sqrt{13}}{3}\). The correct answer is option (3).