Question

The internal center of similitude of two circles $${(x-3)}^2+{(y-2)}^2=9,{(x+5)}^2+{(y+6)}^2=9$$ is
$$\begin{array}{rl}& \text{(A) (-1,-2)}\\ & \text{(B) (-2,-1)}\\ & \text{(C) (3,2)}\\ & \text{(D) (-5,-6)}\end{array}$$

Answer 1

Hint: We know that if $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ are divided by $$R(x_3,y_3)$$ in the ratio $$m:n$$ internally if $$x_3={\displaystyle \frac{m{x}_2+n{x}_1}{m+n}},y_3={\displaystyle \frac{m{y}_2+n{y}_1}{m+n}}$$. We should know that if the internal centre of similitude of two circles $${(x-x_1)}^2+{(y-y_1)}^2=r_1^2,{(x-x_2)}^2+{(y-y_2)}^2=r_2^2$$ is $$A(x_3,y_3)$$, then $$A(x_3,y_3)$$ divides the centre of two circles in the ratio $$r_1:r_2$$. By using this concept, we can find the internal centre of similitude of $${(x-3)}^2+{(y-2)}^2=9,{(x+5)}^2+{(y+6)}^2=9$$.

Complete step by step solution:
We know that if the internal centre of similitude of two circles $${(x-x_1)}^2+{(y-y_1)}^2=r_1^2,{(x-x_2)}^2+{(y-y_2)}^2=r_2^2$$ is $$A(x_3,y_3)$$, then $$A(x_3,y_3)$$ divides the centre of two circles in the ratio $$r_1:r_2$$ .

In the question, we were given two circles $${(x-3)}^2+{(y-2)}^2=9,{(x+5)}^2+{(y+6)}^2=9$$.
Now let us compare $${(x-3)}^2+{(y-2)}^2=9$$ with $${(x-x_1)}^2+{(y-y_1)}^2=r_1^2$$.
We get
$$\begin{array}{rl}& x_1=3....(1)\\ & y_1=2.....(2)\\ & {r_1}^2=9\\ & \Rightarrow r_1=3....(3)\end{array}$$
Now let us compare $${(x+5)}^2+{(y+6)}^2=9$$ with $${(x-x_2)}^2+{(y-y_2)}^2=r_2^2$$.
We get
$$\begin{array}{rl}& x_2=-5...(4)\\ & y_2=-6.....(5)\\ & {r_2}^2=9\\ & \Rightarrow r_2=3....(6)\end{array}$$
We know that the internal centre of similitude divides the line joining the centre in the ratio $$r_1:r_2$$ internally.
So, we get the centre of $${(x-3)}^2+{(y-2)}^2=9$$ is $$C_1(3,2)$$ and radius of $${(x-3)}^2+{(y-2)}^2=9$$ is equal to 3.
In the similar manner, we get centre of $${(x+5)}^2+{(y+6)}^2=9$$ is $$C_2(-5,-6)$$ and radius of $${(x+5)}^2+{(y+6)}^2=9$$ is equal to 3.
So, it is clear that $$A(x_3,y_3)$$ divides $$C_1(3,2)$$ and $$C_2(-5,-6)$$ in the ratio $$3:3$$ internally.
We know that if $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ are divided by $$R(x_3,y_3)$$ in the ratio $$m:n$$ internally if $$x_3={\displaystyle \frac{m{x}_2+n{x}_1}{m+n}},y_3={\displaystyle \frac{m{y}_2+n{y}_1}{m+n}}$$.
By using this concept, we get
$$\begin{array}{rl}& \Rightarrow x_3={\displaystyle \frac{m{x}_2+n{x}_1}{m+n}}\\ & \Rightarrow x_3={\displaystyle \frac{3(-5)+3(3)}{3+3}}\\ & \Rightarrow x_3=-1.....(7)\end{array}$$
In the similar way, we get
$$\begin{array}{rl}& \Rightarrow y_3={\displaystyle \frac{m{y}_2+n{y}_1}{m+n}}\\ & \Rightarrow y_3={\displaystyle \frac{3(-6)+3(2)}{3+3}}\\ & \Rightarrow y_3=-2.....(8)\end{array}$$
From equation (7) and equation (8), we get the coordinates of internal centre of similitude is $$(-1,-2)$$.
Hence, option A is correct.

Note: Some students have a misconception that if $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ are divided by $$R(x_3,y_3)$$ in the ratio $$m:n$$ internally if $$x_3={\displaystyle \frac{m{x}_2-n{x}_1}{m-n}},y_3={\displaystyle \frac{m{y}_2-n{y}_1}{m-n}}$$. But we know that the given coordinates are the coordinates obtained if $$R(x_3,y_3)$$ is get divided in the ratio $$m:n$$ externally. But if this misconception is followed, then we will get the external centre of similitude. But we want internal central similitude. So, this misconception hasto be avoided.