Question

The external centre of similitude of the two circles \[ x^2+y^2-4x+6y+4=0 \] and \[ x^2+y^2-2x+2y-2=0 \] is

• A. \((1,-3)\)
• B. \((-1,3)\)
• C. \((-1,-3)\)
• D. \((1,3)\) \bigskip

Hint:

For two circles with radii \(r_1\) and \(r_2\): \[ \text{External centre of similitude} = \left( \frac{r_1x_2-r_2x_1}{r_1-r_2}, \frac{r_1y_2-r_2y_1}{r_1-r_2} \right). \] First convert each circle into centre-radius form by completing squares.

Answer 1

The Correct Option is B

Concept: The external centre of similitude divides the segment joining the centres externally in the ratio of their radii. Use the section formula directly, then simplify to get the coordinates.

Step 1: Complete squares for both circles. Circle 1: \(x^2+y^2-4x+6y+4=0 \Rightarrow (x-2)^2+(y+3)^2=9\). Centre C1(2,-3), r1=3. Circle 2: \(x^2+y^2-2x+2y-2=0 \Rightarrow (x-1)^2+(y+1)^2=4\). Centre C2(1,-1), r2=2.

Step 2: External division in ratio 3:2. Coordinates: \(x = \frac{3\cdot1 - 2\cdot2}{3-2} = \frac{3-4}{1} = -1\). \(y = \frac{3\cdot(-1) - 2\cdot(-3)}{3-2} = \frac{-3+6}{1} = 3\).

Step 3: The point is \((-1,3)\).

Step 4: Write the final answer. \(\boxed{(-1,3)}\)