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Find the equation of the circle which passes through the points

Question:

Find the equation of the circle which passes through the points (1, 3) and (2, - 1), and has its centre on the line 2x + y – 4 = 0.

 

Solution:

The equation of a circle: $x^{2}+y^{2}+2 g x+2 f y+c=0 \ldots$ (i)

Putting $(1,3) \&(2,-1)$ in (i)

$2 g+6 f+c=-10 . .($ ii $)$

$4 g-2 f+c=-5 . .($ iii $)$

Since the centre lies on the given straight line, $(-g,-f)$ must satisfy the equation as

$-2 g-f-4=0 \ldots$ (iv)

Solving, $f=-1, g=-1.5, c=-1$

The equation is $x^{2}+y^{2}-3 x-2 y-1=0$

 

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