Math, asked by sunnyhimeshboddu2580, 8 months ago

Find the values of a, bif ax2 + bxy+3y2–5x+2y-3=0 represents a circle. Also find the radius and
centre of the circle​

Answers

Answered by MaheswariS
7

\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:}}

\underline{\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}}}}

\underline{\textsf{Find more:}}

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