A circle cuts off positive intercepts 5 and 6 | vedclass

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(C) The general equation of a circle passing through the origin $(0,0)$ is given by $x^2+y^2+2gx+2fy=0$.Since the circle cuts off intercepts on the ax

Vedclass · Accepted Answer

$x^2+y^2-5x-6y=0$

Vedclass · Answer

(C) The general equation of a circle passing through the origin $(0,0)$ is given by $x^2+y^2+2gx+2fy=0$.Since the circle cuts off intercepts on the axes,the points of intersection are $(5,0)$ and $(0,6)$.Substituting $(5,0)$ into the equation: $5^2+0^2+2g(5)+2f(0)=0 \implies 25+10g=0 \implies g = -2.5$.Substituting $(0,6)$ into the equation: $0^2+6^2+2g(0)+2f(6)=0 \implies 36+12f=0 \implies f = -3$.Substituting $g$ and $f$ back into the general equation: $x^2+y^2+2(-2.5)x+2(-3)y=0$.This simplifies to $x^2+y^2-5x-6y=0$.

$A$ circle cuts off positive intercepts $5$ and $6$ on the $x$ and $y$ axes respectively,and passes through the origin. Then the equation of the circle is

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