Question

If A, B are the points of contact of the tangents drawn from the point (-3,1) to the circle \( x^2+y^2-4x+2y-4=0 \), then the equation of the circumcircle of the triangle PAB is

• A. \( x^2+y^2-6x+2y-6=0 \)
• B. \( x^2+y^2-x+7=0 \)
• C. \( x^2+y^2+x-7=0 \)
• D. \( x^2+y^2+6x-2y-6=0 \)

Hint:

Remember the property: The circumcircle of $\Delta PAB$ (where P is external, A/B are tangent points) always has the line segment joining P and the circle's center as its diameter.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

The circumcircle of the triangle formed by an external point P and the points of contact A and B passes through P, A, B, and the center of the circle C. The segment PC is the diameter of this circumcircle.
Step 2: Key Formula or Approach:

Circle on diameter connecting \( P(x_1, y_1) \) and Center \( C(x_2, y_2) \): \( (x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0 \).
Step 3: Detailed Explanation:

External point \( P(-3, 1) \). Circle: \( x^2+y^2-4x+2y-4=0 \). Center \( C = (-g, -f) = (2, -1) \). The circumcircle has PC as its diameter. Equation: \[ (x - (-3))(x - 2) + (y - 1)(y - (-1)) = 0 \] \[ (x + 3)(x - 2) + (y - 1)(y + 1) = 0 \] \[ (x^2 + x - 6) + (y^2 - 1) = 0 \] \[ x^2 + y^2 + x - 7 = 0 \]
Step 4: Final Answer:

The equation is \( x^2 + y^2 + x - 7 = 0 \).