Find the equation of the circle …
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Yogita Ingle 3 years, 4 months ago
Let the equation of circle be
x2+y2+2gx+2fy+c=0 .....(1)
where (−g,−f) is center and g2+f2−c is the radius.
Since, the circle passes through (2,3), so it satisfies eqn (1)
⇒4+9+4g+6f+c=0
⇒4g+6f+c=−13 .....(2)
Since, the circle passes through (−1,1), so it satisfies eqn (1)
⇒1+1−2g+2f+c=0
⇒−2g+2f+c=−2 .....(3)
Subtracting eqn (3) from eqn (2), we get
6g+4f=−11 ....(4)
Given center lies on the line x−3y−11=0
Since, center (−g,−f) satisfies this equation.
⇒−g+3f=11 ....(5)
Solving eqn (4) and (5), we get
g= −7/2 ,f= 5/2
Put this value in (2), we get
c=−14
Substituting these values in (1),
x2+y2−7x+5y−14=0
which is the equation of required circle.
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