Question

The length of the diameter of the circle which touches the $x-$axis at the point $(1,0)$ and passes through the point $(2,3)$

• A. $6/5$
• B. $5/3$
• C. $10/3$
• D. $3/5$.

Answer 1

Answer: (C)

$10/3$

Given:-

A circle which touches the $x$-axxis at $(1,0)$ and also passes through the point $(2,3)$.

To find:-

Diameter of the circle

Solution:-

Let the coordinates of centre of the circle be $(h,k)$ and $r$ be the radius of the circle.

Now,

Since the circle touches $x$-axis at $(1,0)$.

$\therefore$ Radius of circle $\left(r\right)=k$

Therefore, equation of circle is-

${(x-h)}^2+{(y-k)}^2=k^2$

Given that the circle passes through points $(1,0)$ and $(2,3)$, thus both coordinates will satisfy the equation of circle.

Therefore,

${(1-h)}^2+k^2=k^2$

$\Rightarrow {(h-1)}^2=0$

$\Rightarrow h=1.....\left(1\right)$

${(2-h)}^2+{(3-k)}^2=k^2$

$\Rightarrow h^2+4-4h+9+k^2-6k=k^2$

$\Rightarrow h^2-4h-6k+13=0.....\left(2\right)$

From ${eq}^n\left(1\right)$, we have

$1^2-4\left(1\right)-6k+13=0$

$1-4-6k+13=0$

$\Rightarrow 6k=10$

$\Rightarrow k=\frac{5}{3}$

Therefore,

Radius of circle $=\frac{5}{3}$

$\therefore$ Diameter of circle $=2r=2\times \frac{5}{3}=\frac{10}{3}$

Thus the diameter of circle is $\frac{10}{3}$.

Hence the correct answer is $\left(C\right)\frac{10}{3}$