If one of the diameters of the circle, given by the equation ( x^2 + y^2 - 4x + 6y - 12 = 0 ), is a chord of a circle, 'S', whose centre is at ( (-3, 2) ), then the length of radius of 'S' is ________ units.
Show Hint
For a circle where a diameter is a chord of another circle, the distance between centres, radius of the first circle, and radius of the second circle form a right-angled triangle.
Step 1: Understanding the Question:
We have two circles. The diameter of the first circle acts as a chord for the second circle \( S \). Step 3: Detailed Explanation:
1. First circle \( C_1 \): \( x^2 + y^2 - 4x + 6y - 12 = 0 \).
Centre \( O_1 = (2, -3) \), Radius \( r_1 = \sqrt{2^2 + (-3)^2 - (-12)} = \sqrt{4+9+12} = 5 \).
2. The diameter of \( C_1 \) is the chord of \( S \). Length of chord \( L = 2 \times r_1 = 10 \).
3. The midpoint of this chord must be the centre of \( C_1 \), which is \( O_1(2, -3) \).
4. Centre of circle \( S \) is \( O_S = (-3, 2) \).
5. Distance from \( O_S \) to chord midpoint \( O_1 \):
\[ d = \sqrt{(2 - (-3))^2 + (-3 - 2)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2} \]
6. Radius \( R \) of circle \( S \) satisfies:
\[ R^2 = d^2 + (\text{half-chord length})^2 \]
\[ R^2 = (5\sqrt{2})^2 + 5^2 = 50 + 25 = 75 \]
\[ R = \sqrt{75} = 5\sqrt{3} \text{ units} \] Step 4: Final Answer:
The radius of circle \( S \) is \( 5\sqrt{3} \).