Question

Find the equation of circle passing through a point (1,1) (2,-1) and (3,5)
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Please solve fast, no wrong answer otherwise I will report

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{Points are (1,1), (2,-1) and (3,5)}$

$\mathsf{}$

$\underline{\textsf{To find:}}$

$\textsf{The equation of the circle pass through the given points}$

$\underline{\textsf{Solution:}}$

$\mathsf{Let\;the\;equation\;of\;the\;circle\be\;x^2+y^2+2gx+2fy+c=0}$

$\mathsf{Since the circle pass through points (1,1), (2,-1) and (3,5), we have}$

$\mathsf{1^2+1^2+2g(1)+2f(1)+c=0}$

$\mathsf{1+1+2g+2f+c=0}$

$\implies\mathsf{2g+2f+c=-2}$......(1)

$\mathsf{2^2+(-1)^2+2g(2)+2f(-1)+c=0}$

$\mathsf{4+1+4g-2f+c=0}$

$\implies\mathsf{4g-2f+c=-5}$......(2)

$\mathsf{3^2+5^2+2g(3)+2f(5)+c=0}$

$\mathsf{9+25+6g+10f+c=0}$

$\implies\mathsf{6g+10f+c=-34}$......(3)

$\mathsf{(1)-(2)\;\implies}$

$\mathsf{-2g+4f=3}$.........(4)

$\mathsf{(2)-(3)\;\implies}$

$\mathsf{-2g-12f=29}$.......(5)

$\mathsf{16f=32}$

$\implies\boxed{\mathsf{f=2}}$

$\mathsf{Put\;f=2\;in\;(4)}$

$\mathsf{-2g+8=3}$

$\mathsf{-2g=-5}$

$\implies\mathsf{g=\dfrac{5}{2}}$

$\textsf{Substitute the values of f and g in (1)}$

$\mathsf{5+4+c=-2}$

$\implies\boxed{\mathsf{c=-11}}$

$\textsf{Substituting the values of g,f and c}$

$\textsf{The equation of the required circle is}$

$\mathsf{x^2+y^2+2(\dfrac{5}{2})x+2(2)y-11=0}$

$\implies\boxed{\mathsf{x^2+y^2+5x+4y-11=0}}$

$\underrline{\textsf{Find more:}}$

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