A circle passes through the origin and makes | vedclass

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(A) Since the circle passes through the origin $(0, 0)$ and cuts intercepts $a$ and $b$ on the $x$ and $y$ axes respectively,the circle passes through

Vedclass · Accepted Answer

$x^2 + y^2 - ax - by = 0$

Vedclass · Answer

(A) Since the circle passes through the origin $(0, 0)$ and cuts intercepts $a$ and $b$ on the $x$ and $y$ axes respectively,the circle passes through the points $(a, 0)$ and $(0, b)$.Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(0, 0)$,we have $c = 0$.Since it passes through $(a, 0)$,we have $a^2 + 2ga = 0$,which implies $g = -a/2$.Since it passes through $(0, b)$,we have $b^2 + 2fb = 0$,which implies $f = -b/2$.Substituting these values into the general equation,we get $x^2 + y^2 - ax - by = 0$.

$A$ circle passes through the origin and makes intercepts $a$ and $b$ on the coordinate axes. The equation of the circle is:

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