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

We are given a circle with the equation:

\(x^2 + y^2 - 4x - 6y - 3 = 0\) 

First, let's rewrite the equation to find the center and radius of the circle. We complete the square for the \(x\) and \(y\) terms:

The given equation can be rearranged as:

\((x^2 - 4x) + (y^2 - 6y) = 3\)

Complete the squares:

\((x^2 - 4x + 4) + (y^2 - 6y + 9) = 3 + 4 + 9\) \((x - 2)^2 + (y - 3)^2 = 16\)

The center of the circle \(O\) is \((2, 3)\) and the radius is \(4\).

Lines \(PQ\) and \(MN\) are tangents to the circle at points \(A\) and \(B\) respectively, and \(\angle AOB = \pi/3\).

We need to find the locus of the intersection of \(PQ\) and \(MN\).

The angle \(\angle AOB\) implies that the lines subtend an angle of \(\pi/3\) at the center, given.

The locus of intersection of two tangents drawn from an external point that subtends an angle \(\theta\) at the circle center is a circle with the same center but with radius:

\(R \cdot \sec(\theta/2)\)

For \(\theta = \pi/3\), \(\sec(\pi/6) = 2/\sqrt{3}\). Thus, the radius of the locus circle is:

\(4 \cdot \frac{2}{\sqrt{3}} = \frac{8}{\sqrt{3}}\)

Simplifying, this becomes:

\(\frac{8\sqrt{3}}{3}\)

The locus circle equation is therefore:

\((x - 2)^2 + (y - 3)^2 = \left(\frac{8\sqrt{3}}{3}\right)^2\)

Calculating the expansion:

\((x^2 - 4x + 4) + (y^2 - 6y + 9) = \frac{64}{3}\)

Further simplifying gives:

\(x^2 + y^2 - 4x - 6y + 13 = \frac{64}{3}\)

Or multiply throughout by 3 to clear the fraction:

\(3x^2 + 3y^2 - 12x - 18y + 39 = 64\)

Finally simplifying leads to:

\(3x^2 + 3y^2 - 12x - 18y - 25 = 0\)

Therefore, the equation of the locus is:

\(3(x^2 + y^2) - 12x - 18y - 25 = 0\)

Hence, the correct answer is:

\(3(x^2+y^2)-12x-18y-25=0\)