Question

Find the value of a,b if ax^2+bxy+3y^2-5x+2y-3=0 represents a circle Also find radius and centre of the circle

Answer 1

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

$\textsf{Circle is}$

$\mathsf{ax^2+bxy+3y^2-5x+2y-3=0}$

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

$\textsf{(i)The value of a and b }$

$\textsf{(ii)Centre and radius}$

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

$underrline{\textsf{Concept used:}}$

$\textsf{In any equation of circle,}$

$\mathsf{(i)coefficient\;of\;x^2=coefficient\;of\;y^2}$

$\mathsf{(ii)coefficient\;of\;xy=0}$

$\implies\boxed{\mathsf{a=3\;and\;b=0}}$

$\mathsf{Now,\;the\;given\;circle\;can\;written\;as}$

$\mathsf{3x^2+3y^2-5x+2y-3=0}$

$\implies\mathsf{x^2+y^2-\dfrac{5}{3}x+\dfrac{2}{3}y-1=0}$

$\textsf{Comparing this equation with}\:\mathsf{x^2+y^2+2gx+2fy+c=0\;we\;get}$

$\mathsf{2g=\dfrac{-5}{3}\;\;and\;\;2f=\dfrac{2}{3}}$

$\mathsf{g=\dfrac{-5}{6}\;\;and\;\;f=\dfrac{1}{3}}$

$\textsf{Centre is (-g,-f)}$

$\boxed{\mathsf{Centre\;is\;\left(\dfrac{5}{6},\dfrac{-1}{3}\right)}}$

$\mathsf{Radius=\sqrt{g^2+f^2-c}}$

$\mathsf{Radius=\sqrt{\dfrac{25}{36}+\dfrac{1}{9}+1}}$

$\mathsf{Radius=\sqrt{\dfrac{25+4+36}{36}}}$

$\implies\boxed{\mathsf{Radius=\dfrac{\sqrt{65}}{6}}}}$

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