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(B) Since the circle touches the $y$-axis at the origin $(0, 0)$,its center must lie on the $x$-axis. Let the center be $(h, 0)$.Since it touches the

Vedclass · Accepted Answer

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

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(B) Since the circle touches the $y$-axis at the origin $(0, 0)$,its center must lie on the $x$-axis. Let the center be $(h, 0)$.Since it touches the $y$-axis at the origin,the radius of the circle is $|h|$.The equation of the circle is $(x - h)^2 + (y - 0)^2 = h^2$,which simplifies to ${x^2} - 2hx + {h^2} + {y^2} = {h^2}$,or ${x^2} + {y^2} - 2hx = 0$.Since the circle passes through the point $(2, 1)$,we substitute these coordinates into the equation:${2^2} + {1^2} - 2h(2) = 0$$4 + 1 - 4h = 0$$5 - 4h = 0$$h = \frac{5}{4}$Substituting $h = \frac{5}{4}$ back into the equation ${x^2} + {y^2} - 2hx = 0$:${x^2} + {y^2} - 2(\frac{5}{4})x = 0$${x^2} + {y^2} - \frac{5}{2}x = 0$Multiplying by $2$,we get $2{x^2} + 2{y^2} - 5x = 0$.Thus,the correct option is $B$.

The equation of the circle passing through the point $(2, 1)$ and touching the $y$-axis at the origin is:

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