Question

Let PQ and MN be two straight lines touching the circle \(x^2+y^2-4x-6y-3=0\) at the points A and B respectively. Let O be the centre of the circle and \(\angle AOB = \pi/3\). Then the locus of the point of intersection of the lines PQ and MN is:

• A. \(x^2+y^2-18x-12y-25=0\)
• B. \(3(x^2+y^2)-18x-12y+25=0\)
• C. \(3(x^2+y^2)-12x-18y-25=0\)
• D. \(x^2+y^2-12x-18y-25=0\)

Hint:

For locus problems involving tangents to a circle from an external point P, always draw a diagram. The key geometric figure is often the right-angled triangle formed by the center O, point of tangency A, and the external point P. Using simple trigonometry in this triangle is usually the fastest way to find a condition on the distance OP.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We have a circle and two tangents drawn from a point P (the intersection of lines PQ and MN) to the circle. The points of tangency are A and B. We are given the angle subtended by the chord of contact AB at the center O. We need to find the locus of the point P.
Step 2: Key Formula or Approach:
1. Find the center and radius of the given circle. The equation is \(x^2+y^2+2gx+2fy+c=0\), center is \((-g, -f)\) and radius is \(\sqrt{g^2+f^2-c}\).
2. Use the geometry of the situation. Let the point of intersection of tangents be P(h, k). The quadrilateral OAPB is formed by the center O, the points of tangency A and B, and the external point P. 3. In the quadrilateral OAPB, OA and OB are radii and are perpendicular to the tangents PA and PB. Therefore, \(\angle OAP = \angle OBP = 90^\circ\).
4. The triangle \(\triangle OAP\) is a right-angled triangle. We can use trigonometry to relate the distance OP to the radius and the angle \(\angle AOP\).
5. The line segment OP bisects the angle \(\angle AOB\). So, \(\angle AOP = \frac{1}{2} \angle AOB\).
Step 3: Detailed Explanation:
The equation of the circle is \(x^2+y^2-4x-6y-3=0\).
The center of the circle is O\(( -(-4/2), -(-6/2) ) = (2, 3)\).
The radius is \(r = \sqrt{(-2)^2 + (-3)^2 - (-3)} = \sqrt{4+9+3} = \sqrt{16} = 4\).
Let the point of intersection of the tangents be P(h, k). Consider the right-angled triangle \(\triangle OAP\) (right-angled at A). We are given \(\angle AOB = \pi/3 = 60^\circ\). The line OP is the angle bisector of \(\angle AOB\), so \(\angle AOP = \frac{1}{2} \angle AOB = \frac{60^\circ}{2} = 30^\circ\). In \(\triangle OAP\), we can use trigonometry: \[ \cos(\angle AOP) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{OA}{OP} \] \[ \cos(30^\circ) = \frac{r}{OP} \] \[ \frac{\sqrt{3}}{2} = \frac{4}{OP} \] \[ OP = \frac{8}{\sqrt{3}} \] The distance of the point P(h,k) from the center O(2,3) is fixed. So, the locus is a circle centered at O. The condition is \(OP^2 = \left(\frac{8}{\sqrt{3}}\right)^2 = \frac{64}{3}\). Using the distance formula for OP: \[ (h-2)^2 + (k-3)^2 = \frac{64}{3} \] To get the equation of the locus, we replace (h, k) with (x, y): \[ (x-2)^2 + (y-3)^2 = \frac{64}{3} \] Expand the equation: \[ 3[(x^2 - 4x + 4) + (y^2 - 6y + 9)] = 64 \] \[ 3(x^2 + y^2 - 4x - 6y + 13) = 64 \] \[ 3x^2 + 3y^2 - 12x - 18y + 39 = 64 \] \[ 3x^2 + 3y^2 - 12x - 18y - 25 = 0 \] \[ 3(x^2 + y^2) - 12x - 18y - 25 = 0 \] Step 4: Final Answer:
The locus of the point of intersection is \(3(x^2+y^2)-12x-18y-25=0\).