Question

If one end of a diameter of the circle $x^2+y^2-4x-6y+11=0$ be $(3,4)$, then the other end is

• A. $(0,0)$
• B. $(1,1)$
• C. $(1,2)$
• D. $(2,1)$

Answer 1

The correct option is C $(1,2)$

Equation of the given circle is $x^2+y^2-4x-6y+11=0$ is of the form $x^2+y^2+2gx+2fy+c=0$

where $2g=-4\Rightarrow g=-2$

$2f=-6\Rightarrow f=-3$

Hence the centre of the circle is $(2,3)$

The coordinate of the point which lies on the circle is $(3,4)$

Let the coordinates of the point on the other end of the diameter be $(x_1,y_1)$.

The coordinate of midpoint of these two points is $(2,3)$

$\frac{x_1+3}{2}=2,\frac{y_1+4}{2}=3$

$\Rightarrow x_1+3=4,y_1+4=6$

$\Rightarrow x_1=4-3=1,y_1=6-4=2$

$\therefore x_1=1,y_1=2$

Hence the coordinates of the point on the other end of the diameter is $(1,2)$