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
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
Show 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.
Updated On: Jul 9, 2026
- \((1,-3)\)
- \((-1,3)\)
- \((-1,-3)\)
- \((1,3)\) \bigskip
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
If two circles have centres
\[
C_1(x_1,y_1), \qquad C_2(x_2,y_2)
\]
and radii
\[
r_1,\qquad r_2,
\]
then their external centre of similitude divides the line joining the centres externally in the ratio
\[
r_1:r_2.
\]
The coordinates are given by
\[
\left(
\frac{r_1x_2-r_2x_1}{r_1-r_2},
\frac{r_1y_2-r_2y_1}{r_1-r_2}
\right).
\]
Step 1: Find the centre and radius of the first circle. Given \[ x^2+y^2-4x+6y+4=0. \] Completing squares, \[ (x-2)^2+(y+3)^2=9. \] Hence, \[ C_1=(2,-3), \qquad r_1=3. \]
Step 2: Find the centre and radius of the second circle. Given \[ x^2+y^2-2x+2y-2=0. \] Completing squares, \[ (x-1)^2+(y+1)^2=4. \] Hence, \[ C_2=(1,-1), \qquad r_2=2. \]
Step 3: Apply the external division formula. The external centre of similitude divides \[ C_1C_2 \] externally in the ratio \[ 3:2. \] Therefore, \[ x = \frac{3(1)-2(2)}{3-2} = \frac{3-4}{1} = -1. \] \[ y = \frac{3(-1)-2(-3)}{3-2} = \frac{-3+6}{1} = 3. \]
Step 4: Write the final answer. Hence the external centre of similitude is \[ \boxed{(-1,3)}. \]
Step 1: Find the centre and radius of the first circle. Given \[ x^2+y^2-4x+6y+4=0. \] Completing squares, \[ (x-2)^2+(y+3)^2=9. \] Hence, \[ C_1=(2,-3), \qquad r_1=3. \]
Step 2: Find the centre and radius of the second circle. Given \[ x^2+y^2-2x+2y-2=0. \] Completing squares, \[ (x-1)^2+(y+1)^2=4. \] Hence, \[ C_2=(1,-1), \qquad r_2=2. \]
Step 3: Apply the external division formula. The external centre of similitude divides \[ C_1C_2 \] externally in the ratio \[ 3:2. \] Therefore, \[ x = \frac{3(1)-2(2)}{3-2} = \frac{3-4}{1} = -1. \] \[ y = \frac{3(-1)-2(-3)}{3-2} = \frac{-3+6}{1} = 3. \]
Step 4: Write the final answer. Hence the external centre of similitude is \[ \boxed{(-1,3)}. \]
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