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Question

Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x+y=16

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Solution

Let the equation of circle be
x2+y2+2gx+2fy+c=0 .....(1)
where (g,f) is center and g2+f2c is the radius.
Since, the circle passes through (4,1), so it satisfies eqn (1)
16+1+8g+2f+c=0
8g+2f+c=17 .....(2)
Since, the circle passes through (6,5), so it satisfies eqn (1)
36+25+12g+10f+c=0
12g+10f+c=61 .....(3)
Subtracting eqn (3) from eqn (2), we get
4g8f=44
g+2f=11 ....(4)
Given center lies on the line 4x+y=16
Since, center (g,f) satisfies this equation.
4gf=16 ....(5)
Solving eqn (4) and (5), we get
g=3,f=4
Put this value in (2), we get
c=15
Substituting these values in (1),
x2+y26x8y+15=0
which is the equation of required circle.

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