Question

The external centre of similitude for circles \[ x^2+y^2+10x-16y-11=0 \] and \[ x^2+y^2-2x+4y-4=0 \] is

• A. \(\left(\frac57,-\frac47\right)\)
• B. \((-2,3)\)
• C. \(\left(\frac{25}{7},-\frac{44}{7}\right)\)
• D. \((-3,5)\)

Hint:

For center of similitude, first convert circles into center-radius form and use section formula carefully.

Answer 1

The Correct Option is C

Step 1: Recall the rule.
The external centre of similitude of two circles divides the line joining their centres externally in the ratio of their radii $r_1:r_2$.
Step 2: First circle.
For $x^2+y^2+10x-16y-11=0$, we have $g=5$, $f=-8$, $c=-11$, so the centre is $C_1=(-5,8)$ and $r_1=\sqrt{25+64+11}=\sqrt{100}=10$.
Step 3: Second circle.
For $x^2+y^2-2x+4y-4=0$, $g=-1$, $f=2$, $c=-4$, so the centre is $C_2=(1,-2)$ and $r_2=\sqrt{1+4+4}=3$.
Step 4: External division formula.
A point dividing $C_1C_2$ externally in ratio $r_1:r_2=10:3$ has coordinates \[ \left(\frac{r_1 x_2-r_2 x_1}{r_1-r_2},\ \frac{r_1 y_2-r_2 y_1}{r_1-r_2}\right). \]
Step 5: Substitute.
The $x$-coordinate is $\dfrac{10(1)-3(-5)}{10-3}=\dfrac{10+15}{7}=\dfrac{25}{7}$, and the $y$-coordinate is $\dfrac{10(-2)-3(8)}{7}=\dfrac{-20-24}{7}=-\dfrac{44}{7}$.
Step 6: Box the answer.
The external centre of similitude is $\left(\tfrac{25}{7},-\tfrac{44}{7}\right)$, option (3).
\[ \boxed{\left(\tfrac{25}{7},-\tfrac{44}{7}\right)\ \text{(option 3)}} \]