KCET · Maths · Circle
Equation of the circle centred at \((4,3)\) touching the circle \(x^{2}+y^{2}=1\) externally, is
- A \(x^{2}+y^{2}-8 x-6 y+9=0\)
- B \(x^{2}+y^{2}+8 x+6 y+9=0\)
- C \(x^{2}+y^{2}+8 x-6 y+9=0\)
- D \(x^{2}+y^{2}-8 x+6 y+9=0\)
Answer & Solution
Correct Answer
(A) \(x^{2}+y^{2}-8 x-6 y+9=0\)
Step-by-step Solution
Detailed explanation
Given that, equation of circle
\[
x^{2}+y^{2}=1
\]
Centre at \(\mathrm{O} \rightarrow(0,0)\)
Radius \(=\mathrm{OA}=1\)
Also, the centre of another circle \(\rightarrow C(4,3)\) both circle touch externally. Then, distance between centres \(=\) OC.
\[
=\sqrt{(4-0)^{2}+(3-0)^{2}}=\sqrt{16-9}=5
\]
Now,
\(\mathrm{AC}=\mathrm{OC}-\mathrm{OA}\)
\[
A C=5-1=4
\]
So, the radius of other circle is 4 .
Now, the equation of other circle touch externally to the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=1\) is,
\[
\begin{gathered}
(x-4)^{2}+(y-3)^{2}=16 \\
\Rightarrow \quad x^{2}+y^{2}-8 x-6 y+9=0
\end{gathered}
\]
\[
x^{2}+y^{2}=1
\]
Centre at \(\mathrm{O} \rightarrow(0,0)\)
Radius \(=\mathrm{OA}=1\)
Also, the centre of another circle \(\rightarrow C(4,3)\) both circle touch externally. Then, distance between centres \(=\) OC.
\[
=\sqrt{(4-0)^{2}+(3-0)^{2}}=\sqrt{16-9}=5
\]
Now,
\(\mathrm{AC}=\mathrm{OC}-\mathrm{OA}\)
\[
A C=5-1=4
\]
So, the radius of other circle is 4 .
Now, the equation of other circle touch externally to the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=1\) is,
\[
\begin{gathered}
(x-4)^{2}+(y-3)^{2}=16 \\
\Rightarrow \quad x^{2}+y^{2}-8 x-6 y+9=0
\end{gathered}
\]
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