Question

The point of intersection of the lines \[ 2x+3y-12=0 \] and \[ 3x-2y-5=0 \] is the centre of a circle \(S=0\). If \(AB\) is a chord of \(S=0\) and it is a diameter of the circle \[ x^2+y^2-10x+4y+13=0, \] then the radius of the circle \(S=0\) is

• A. \(4\)
• B. \(6\)
• C. \(16\)
• D. \(12\) \bigskip

Hint:

If a chord of one circle is a diameter of another circle, first compute the diameter length. Then use \[ \text{Chord Length} = 2\sqrt{R^2-d^2} \] to determine the unknown radius.

Answer 1

The Correct Option is B

Concept: The centre of circle S lies at the intersection of its diameter lines. The given circle has a known centre and radius. The chord AB of S is a diameter of the given circle, so its length is 8. Use chord length formula for S to find its radius.

Step 1: Find centre C of S. Solve \(2x+3y-12=0\) and \(3x-2y-5=0\): add 2×(first) + 3×(second)? Better: multiply first by 2, second by 3: \(4x+6y=24, 9x-6y=15\). Sum: \(13x=39 \Rightarrow x=3, y=2\). C(3,2).

Step 2: Given circle: \(x^2+y^2-10x+4y+13=0 \Rightarrow (x-5)^2+(y+2)^2=16\). Centre C1(5,-2), radius 4.

Step 3: Distance CC1 = \(\sqrt{(5-3)^2+(-2-2)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt5\).

Step 4: AB is a chord of S and a diameter of the given circle, so AB = 8. The perpendicular from C to AB meets AB at its midpoint, which is C1. So distance from C to AB is CC1 = 2√5.

Step 5: For circle S: \(\frac12 \text{chord} = \sqrt{R^2 - d^2} \Rightarrow 4 = \sqrt{R^2 - (2\sqrt5)^2} \Rightarrow 16 = R^2 - 20 \Rightarrow R^2 = 36 \Rightarrow R=6\).

Step 6: Write the final answer. \(\boxed{6}\)