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Find the centre of the circle passing through $(5, -8), (2, -9)$ and $(2, 1)$.
Given:
The centre of a circle passes through $(5, -8), (2, -9)$ and $(2, 1)$.
To do:
We have to find the centre of the given circle.
Solution:
Let \( \mathrm{O} \) is the centre of the circle and \( \mathrm{A}(5,-8), \mathrm{B} \) (2,-9) and \( \mathrm{C}(2,1) \) are the points on the circle.
Let the coordinates of \( \mathrm{O} \) are \( (x, y) \).
This implies,
\( \mathrm{OA}=\mathrm{OB}=\mathrm{OC} \) (Radii of the circle)
\( \mathrm{OA}^{2}=\mathrm{OB}^{2}=\mathrm{OC}^{2} \)
On squaring, we get,
\( \mathrm{OA}^{2}=(x-5)^{2}+(y+8)^{2} \)
\( =x^{2}-10 x+25+y^{2}+16 y+64 \)
\( =x^{2}+y^{2}-10 x+16 y+89 \)
\( \mathrm{OB}^{2}=(x-2)^{2}+(y+9)^{2} \)
\( =x^{2}+4-4 x+y^{2}+81+18 y \)
\( =x^{2}+y^{2}-4 x+18 y+85 \)
\( \mathrm{OC}^{2}=(x-2)^{2}+(y-1)^{2} \)
\( =x^{2}-4 x+4+y^{2}-2 y+1 \)
\( =x^{2}+y^{2}-4 x-2 y+5 \)
\( \mathrm{OA}^{2}=\mathrm{OB}^{2} \)
\( \Rightarrow x^{2}+y^{2}-10 x+16 y+89=x^{2}+y^{2}-4 x+18 y+85 \)
\( \Rightarrow -10 x+4 x+16 y-18 y=85-89 \)
\( \Rightarrow -6 x-2 y=-4 \)
\( \Rightarrow -2(3 x+y)=-2(2) \)
\( \Rightarrow 3 x+y=2 \).........(i)
\( \mathrm{OB}^{2}=\mathrm{OC}^{2} \)
\( \Rightarrow x^{2}+y^{2}-4 x+18 y+85=x^{2}+y^{2}-4 x-2 y+5 \)
\( \Rightarrow 18 y+2 y=5-85 \)
\( \Rightarrow 20 y=-80 \)
\( \Rightarrow y=\frac{-80}{20}=-4 \)
Substituting the value of \( y \) in (i), we get,
\( \Rightarrow 3 x-4=2 \)
\( \Rightarrow 3 x=2+4=6 \)
\( \Rightarrow x=\frac{6}{3}=2 \)
Therefore, the centre of the given circle is $(2, -4)$.
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