Question

If one end of a diameter of the circle $3x^2+3y^2-9x+6y+5=0$ is $(1,2)$, then the other end is

• A. $(2,1)$
• B. $(2,4)$
• C. $(2,-4)$
• D. $(-4,2)$

Answer 1

The correct option is C $(2,-4)$

Given equation of circle is
$3x^2+3y^2-9x+6y+5=0$
$\Rightarrow$ $x^2+y^2-3x+2y+\frac{5}{3}=0$
Centre $=(\frac{3}{2},-1)$
and radius $=\sqrt{\frac{9}{4}+1-\frac{5}{3}}$
$=\sqrt{\frac{19}{12}}=\frac{1}{2}\sqrt{\frac{19}{3}}$
We know that centre of the circle is the mid-point of the diameter leaves one end of point of diameter in $(1,2)$.

Let the other end point of diameter is $(h,k)$

Then, $(\frac{3}{2},-1)=(\frac{1+h}{2},\frac{2+k}{2})$
$\Rightarrow$ $\frac{1+h}{2}=\frac{3}{2}$
$\Rightarrow$ $1+h=3$
$\Rightarrow$ $h=2$
and $\frac{2+k}{2}=-1$
$\Rightarrow$ $2+k=-2$
$\Rightarrow$ $k=-4$
So, the other end point is $(2,-4)$