Answer
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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 \[{{x}_{3}}=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},{{y}_{3}}=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\]. 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
\[\begin{align}
& {{x}_{1}}=3....(1) \\
& {{y}_{1}}=2.....(2) \\
& {{r}_{1}}^{2}=9 \\
& \Rightarrow {{r}_{1}}=3....(3) \\
\end{align}\]
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
\[\begin{align}
& {{x}_{2}}=-5...(4) \\
& {{y}_{2}}=-6.....(5) \\
& {{r}_{2}}^{2}=9 \\
& \Rightarrow {{r}_{2}}=3....(6) \\
\end{align}\]
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 \[{{x}_{3}}=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},{{y}_{3}}=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\].
By using this concept, we get
\[\begin{align}
& \Rightarrow {{x}_{3}}=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} \\
& \Rightarrow {{x}_{3}}=\dfrac{3(-5)+3(3)}{3+3} \\
& \Rightarrow {{x}_{3}}=-1.....(7) \\
\end{align}\]
In the similar way, we get
\[\begin{align}
& \Rightarrow {{y}_{3}}=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \\
& \Rightarrow {{y}_{3}}=\dfrac{3(-6)+3(2)}{3+3} \\
& \Rightarrow {{y}_{3}}=-2.....(8) \\
\end{align}\]
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 \[{{x}_{3}}=\dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n},{{y}_{3}}=\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n}\]. 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.
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
\[\begin{align}
& {{x}_{1}}=3....(1) \\
& {{y}_{1}}=2.....(2) \\
& {{r}_{1}}^{2}=9 \\
& \Rightarrow {{r}_{1}}=3....(3) \\
\end{align}\]
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
\[\begin{align}
& {{x}_{2}}=-5...(4) \\
& {{y}_{2}}=-6.....(5) \\
& {{r}_{2}}^{2}=9 \\
& \Rightarrow {{r}_{2}}=3....(6) \\
\end{align}\]
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 \[{{x}_{3}}=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},{{y}_{3}}=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\].
By using this concept, we get
\[\begin{align}
& \Rightarrow {{x}_{3}}=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n} \\
& \Rightarrow {{x}_{3}}=\dfrac{3(-5)+3(3)}{3+3} \\
& \Rightarrow {{x}_{3}}=-1.....(7) \\
\end{align}\]
In the similar way, we get
\[\begin{align}
& \Rightarrow {{y}_{3}}=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \\
& \Rightarrow {{y}_{3}}=\dfrac{3(-6)+3(2)}{3+3} \\
& \Rightarrow {{y}_{3}}=-2.....(8) \\
\end{align}\]
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 \[{{x}_{3}}=\dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n},{{y}_{3}}=\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n}\]. 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.
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