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Question

(A) circle passing through the origin $(0,0)$ and making intercepts of length $a = 6$ on the $x$-axis and $b = 4$ on the $y$-axis has its center at $(

Vedclass · Accepted Answer

$x^2 + y^2 - 6x - 4y = 0$

Vedclass · Answer

(A) circle passing through the origin $(0,0)$ and making intercepts of length $a = 6$ on the $x$-axis and $b = 4$ on the $y$-axis has its center at $(\frac{a}{2}, \frac{b}{2})$.Since the center lies in the first quadrant,the coordinates are $(\frac{6}{2}, \frac{4}{2}) = (3, 2)$.The general equation of a circle passing through the origin with center $(h, k)$ is $x^2 + y^2 - 2hx - 2ky = 0$.Substituting $h = 3$ and $k = 2$,we get $x^2 + y^2 - 2(3)x - 2(2)y = 0$.Thus,the required equation is $x^2 + y^2 - 6x - 4y = 0$.

The equation of the circle passing through the origin,whose center lies in the first quadrant and which makes intercepts of length $6$ and $4$ on the $x$-axis and $y$-axis respectively,is:

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