Find the maximum distance of the point \(K(10,7)\) from the circle \(x^2+y^2-4 x-2 y-20=0\)

Find the maximum distance of the point \(K(10,7)\) from the circle \(x^2+y^2-4 x-2 y-20=0\)
  1. 25
  2. 10
  3. 15
  4. 5

Solution

Equation of given circle is \(\begin{aligned} & x^2+y^2-4 x-2 y-20 & =0 \\ \Rightarrow & (x-2)^2+(y-1)^2 & =25 \end{aligned}\) Having centre \(C(2,1)\) and radius \(r=5\) \(\therefore\) The maximum distance of the point \(K(10,7)\) from the given circle is \(C K+r\) \(\begin{aligned} & =\sqrt{(10-2)^2+(7-1)^2}+5 \\ & =\sqrt{64+36}+5=10+5=15 \text { unit } \end{aligned}\) Hence, option (c) is correct.

Asked in: AP EAMCET 2020 (21 Sep Shift 1)

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