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Circle

Section: Mathematics

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Question : 24 of 89

Marks: +1, -0

If

\left(\;\frac{1}{10},\;\frac{-1}{5}\right)

is the inverse point of a point

(-1,2)

with respect to the circle

x^2+y^2-2 x+4 y+c=0

c=

[AP EAPCET 22 May 2025 S2]

4
-4
2
-2

Solution:

The inverse point is

\left(\;\frac{1}{10},\;\frac{-1}{5}\right)

The original point is

(-1,2)

The circle equation is

x^2+y^2-2 x+4 y+c=0

Centre of the circle is

(1,-2)

The original point is

(-1,2)

\;\text{ The distance }\;\;=\sqrt{(-1-1)^2+(2-(-2))^2}

\;=\sqrt{4+16}=\sqrt{20}

The centre of the circle is

(1,-2)

The inverse point is

\left(\;\frac{1}{10},\;\frac{-1}{5}\right)

\;\;\text{ The distance }\;=\sqrt{ \left(\;\frac{1}{10}-1\right)^2+ \left(\;\frac{-1}{5}+2\right)^2 }

\;=\;\frac{9\sqrt{5}}{10}

\;\Rightarrow \;\; r^2=d_1 d_2=\sqrt{20} \times \;\frac{9\sqrt{5}}{10}=9

\;\Rightarrow \;\; 5-c=9 \Rightarrow c=-4

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