Question

The other end of the diameter through the point $\left(-1,1\right)$ on the circle $x^2+y^2-6x+4y-12=0$ is

• A. $\left(-7,5\right)$
• B. $\left(-7,-5\right)$
• C. $\left(7,-5\right)$
• D. $\left(7,5\right)$

Answer 1

The correct option is C

$\left(7,-5\right)$

Explanation for the correct option

The equation of the circle is $x^2+y^2-6x+4y-12=0$

Now comparing with the general equation of the circle $x^2+y^2+2fx+2ghy+c=0$,where center of circle is $(-g,-f)$ and radius is $\sqrt{g^2+f^2-c}$.

Centre of the circle is $\left(3,-2\right)$.

One end of the diameter is $\left(-1,1\right)$

Let the other end of the diameter is $\left(\alpha ,\beta \right)$

Using mid point formula as we know centre is the mid point of diameter.

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$$\begin{gathered} \therefore \frac{\alpha -1}{2}=3,\frac{\beta +1}{2}=-2 \\ \Rightarrow \alpha =7,\beta =-5 \end{gathered}$$

Therefore the other end of the diameter is $\left(7,-5\right)$

Hence, option (C) is the correct answer.