If one end of the diameter of the circle x^2+ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation of the circle is $x^2+y^2-4x-6y+11=0$.Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $2g=-4 \Rightarrow g=-2

Vedclass · Accepted Answer

$(1,2)$

Vedclass · Answer

(C) The given equation of the circle is $x^2+y^2-4x-6y+11=0$.Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $2g=-4 \Rightarrow g=-2$ and $2f=-6 \Rightarrow f=-3$.The center of the circle is $(-g, -f) = (2, 3)$.Let the other end of the diameter be $(h, k)$.Since the center of the circle is the midpoint of the diameter,we have:$\frac{h+3}{2} = 2$ $\Rightarrow h+3 = 4$ $\Rightarrow h = 1$$\frac{k+4}{2} = 3$ $\Rightarrow k+4 = 6$ $\Rightarrow k = 2$Thus,the other end of the diameter is $(1, 2)$.

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

Similar Questions

Find a possible equation of a chord of the circle $x^2 + y^2 = 100$ which passes through the point $(1, 7)$ and makes an angle of $\frac{2\pi}{3}$ with the origin.

The length of the tangent drawn from the point $(1, 5)$ to the circle $2x^2 + 2y^2 = 3$ is ...

The circle$(s)$ touching the $x$-axis at a distance of $3$ from the origin and having an intercept of length $2 \sqrt{7}$ on the $y$-axis is (are):

If a line passing through the point $\left( -\sqrt{8}, \sqrt{8} \right)$ and making an angle of $135^{\circ}$ with the $x$-axis intersects the circle $x = 5 \cos \theta, y = 5 \sin \theta$ at points $A$ and $B$,then find the length of the chord $AB$.

The points $(x_1, y_1), (x_2, y_2), (x_1, y_2),$ and $(x_2, y_1)$ are always:

Vedclass Test Series

Exam Paper Generator

Online Exam Module