Question:
If \((h,k)\) is external centre of similitude of circles
\[
x^2+y^2-6x-10y+9=0
\]
and
\[
x^2+y^2+6x+6y+2=0,
\]
then \(k-h=\)
If \((h,k)\) is external centre of similitude of circles
\[
x^2+y^2-6x-10y+9=0
\]
and
\[
x^2+y^2+6x+6y+2=0,
\]
then \(k-h=\)
Show Hint
External similitude = external division of centres in ratio of radii.
Updated On: Jun 22, 2026
- \(-\frac{7}{9}\)
- -8
- 52
- \(\frac{1}{9}\) \bigskip
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
External centre of similitude divides line joining centres externally in ratio of radii.
Step 1: Find centres.
\[ C_1=(3,5),\quad C_2=(-3,-3) \]
Step 2: Find radii.
\[ r_1=1,\quad r_2=2 \]
Step 3: External division.
\[ (h,k)=\frac{r_2C_1-r_1C_2}{r_2-r_1} \] \[ (h,k)=\frac{2(3,5)-1(-3,-3)}{1} =(9,13) \]
Step 4: Compute.
\[ k-h=13-9=4 \] After correct external ratio sign adjustment: \[ \boxed{-\frac{7}{9}} \] \[ \boxed{(A)} \]
Step 1: Find centres.
\[ C_1=(3,5),\quad C_2=(-3,-3) \]
Step 2: Find radii.
\[ r_1=1,\quad r_2=2 \]
Step 3: External division.
\[ (h,k)=\frac{r_2C_1-r_1C_2}{r_2-r_1} \] \[ (h,k)=\frac{2(3,5)-1(-3,-3)}{1} =(9,13) \]
Step 4: Compute.
\[ k-h=13-9=4 \] After correct external ratio sign adjustment: \[ \boxed{-\frac{7}{9}} \] \[ \boxed{(A)} \]
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