KCET · Maths · Circle
The least and the greatest distances of the point \((10,7)\) from the circle
\[
x^{2}+y^{2}-4 x-2 y-20=0 \text { are }
\]
- A 10,5
- B 15,20
- C 12,16
- D 5,15
Answer & Solution
Correct Answer
(D) 5,15
Step-by-step Solution
Detailed explanation
Given, \(x^{2}+y^{2}-4 x-2 y-20=0\)
Here, \(g=-2, f=-1, c=-20\)
\[
\therefore \text { Centre }=(-g,-f)=(2,1)
\]
Radius
\[
\begin{aligned}
&=\sqrt{g^{2}+f^{2}-c}=\sqrt{(-2)^{2}+(-1)^{2}-(-20)} \\
&=\sqrt{4+1+20} \\
&=\sqrt{25}=5
\end{aligned}
\]
i.e., The distance between the points \((2,1)\) and
\[
\begin{aligned}
(10,7) &=\sqrt{(10-2)^{2}+(7-1)^{2}} \\
&=\sqrt{64+36} \\
&=\sqrt{100}=10 \text { units }
\end{aligned}
\]
\(\therefore\) The least distance of the point \((10,7)\) from the circle \(=10-5=5\) units and the greatest distance of the point \((10,7)\) from the circle \(=10+5=15\) units.
Here, \(g=-2, f=-1, c=-20\)
\[
\therefore \text { Centre }=(-g,-f)=(2,1)
\]
Radius
\[
\begin{aligned}
&=\sqrt{g^{2}+f^{2}-c}=\sqrt{(-2)^{2}+(-1)^{2}-(-20)} \\
&=\sqrt{4+1+20} \\
&=\sqrt{25}=5
\end{aligned}
\]
i.e., The distance between the points \((2,1)\) and
\[
\begin{aligned}
(10,7) &=\sqrt{(10-2)^{2}+(7-1)^{2}} \\
&=\sqrt{64+36} \\
&=\sqrt{100}=10 \text { units }
\end{aligned}
\]
\(\therefore\) The least distance of the point \((10,7)\) from the circle \(=10-5=5\) units and the greatest distance of the point \((10,7)\) from the circle \(=10+5=15\) units.
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