Let C be a circle having centre in the first | vedclass

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(C) Let the centre of the circle be $(3, r)$ and the radius be $r$, since it touches the $x$-axis at $(3, 0)$.The equation of the circle is $(x - 3)^2

Vedclass · Accepted Answer

$6\sqrt{2}$

Vedclass · Answer

(C) Let the centre of the circle be $(3, r)$ and the radius be $r$, since it touches the $x$-axis at $(3, 0)$.The equation of the circle is $(x - 3)^2 + (y - r)^2 = r^2$.For the $y$-intercept, set $x = 0$: $(0 - 3)^2 + (y - r)^2 = r^2 \Rightarrow 9 + (y - r)^2 = r^2 \Rightarrow (y - r)^2 = r^2 - 9$.Thus, $y = r \pm \sqrt{r^2 - 9}$. The length of the intercept is $2\sqrt{r^2 - 9} = 6\sqrt{3}$.Squaring both sides: $4(r^2 - 9) = 36 	imes 3 = 108 \Rightarrow r^2 - 9 = 27 \Rightarrow r^2 = 36 \Rightarrow r = 6$.The equation of the circle is $(x - 3)^2 + (y - 6)^2 = 36$.The distance $d$ from the centre $(3, 6)$ to the line $x - y - 3 = 0$ is $d = \frac{|3 - 6 - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|-6|}{\sqrt{2}} = 3\sqrt{2}$.The length of the chord is $2\sqrt{r^2 - d^2} = 2\sqrt{36 - (3\sqrt{2})^2} = 2\sqrt{36 - 18} = 2\sqrt{18} = 6\sqrt{2}$.

Let $C$ be a circle having centre in the first quadrant and touching the $x$-axis at a distance of $3$ units from the origin. If the circle $C$ has an intercept of length $6\sqrt{3}$ on the $y$-axis, then the length of the chord of the circle on the line $x - y = 3$ is:

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