Question

The centre of a circle is $(-6,4)$. If one end of the diameter of the circle is at $(-12,8)$, then the other end is at

• A. $(-18,12)$
• B. $(-9,6)$
• C. $(-3,2)$
• D. $(0,0)$

Answer 1

Answer: (D)

$(0,0)$

The centre is the midpont of the line joining the two points of the diameter.

Using Midpoint-formula $(\alpha ,\beta )\equiv (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

From above we have $\alpha =-6,\beta =4$

Let $x_1=x,y_1=y$

Given$x_2=-12,y_2=8$

We have $\alpha =\frac{x_1+x_2}{2}$ and $\beta =\frac{y_1+y_2}{2}$

$x+\frac{(-12)}{2}=-6$

$\Rightarrow x-6=-6$

$\therefore x=0$

$\Rightarrow y+\frac{8}{2}=4$

$\Rightarrow y+4=4$

$\therefore y=0$

Hence, the other point is at $(0,0)$