Let C be a circle having centre in the first quadrant and touching the x-axis at a distanc…

Question

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 $y$-axis, then the length of the chord of the circle C on...

Accepted Answer

Since the circle touches the $x$-axis at a point which is $3$ units from the origin, the point of contact is $(3,0)$.

The centre lies in the first quadrant and the radius is perpendicular to the tangent at the point of contact. So, the centre must be of the form $(3,r)$, where $r$ is the radius.

The circle has an intercept of length $6\sqrt{3}$ on the $y$-axis. This means the chord cut by the line $x=0$ has length $6\sqrt{3}$.

For a circle of radius $r$, if the distance of a chord from the centre is $d$, then chord length is $2\sqrt{{r}^{2}-{d}^{2}}$.

Here, the distance from centre $(3,r)$ to the $y$-axis $(x=0)$ is $d=3$. So the intercept on the $y$-axis is

$2\sqrt{{r}^{2}-{3}^{2}}=6\sqrt{3}$

So,

$\sqrt{{r}^{2}-9}=3\sqrt{3}\Rightarrow {r}^{2}-9=27\Rightarrow {r}^{2}=36\Rightarrow r=6$

Hence, centre of the circle is $(3,6)$ and radius is $6$ units.

Now we need the length of the chord cut by the line $x-y=3$.

First find the perpendicular distance $p$ from the centre $(3,6)$ to the line $x-y-3=0$:

$p=\frac{|1\cdot 3+(-1)\cdot 6-3|}{\sqrt{{1}^{2}+(-1{)}^{2}}}=\frac{|3-6-3|}{\sqrt{2}}=\frac{6}{\sqrt{2}}$

Now chord length on this line is

$PQ=2\sqrt{{6}^{2}-{p}^{2}}\, 	herefore p=\frac{6}{\sqrt{2}}$

$\begin{gathered} p=\frac{|3-6-3|}{\sqrt{2}} \ PQ=2\sqrt{36-\frac{36}{2}} \ =2	imes \frac{6}{\sqrt{2}}=6\sqrt{2} \end{gathered}$

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