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

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

Question

(A) The given equation of the circle is $x^2 + y^2 - 3x + 8y - 4 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g

Vedclass · Accepted Answer

$(-3, -5)$

Vedclass · Answer

(A) The given equation of the circle is $x^2 + y^2 - 3x + 8y - 4 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -3 \Rightarrow g = -\frac{3}{2}$ and $2f = 8 \Rightarrow f = 4$.The center of the circle is $(-g, -f) = (\frac{3}{2}, -4)$.Let the other end of the diameter be $(h, k)$.Since the center is the midpoint of the diameter,we have:$\frac{6 + h}{2} = \frac{3}{2}$ $\Rightarrow 6 + h = 3$ $\Rightarrow h = -3$.$\frac{-3 + k}{2} = -4$ $\Rightarrow -3 + k = -8$ $\Rightarrow k = -5$.Thus,the other end of the diameter is $(-3, -5)$.

If one end of a diameter of the circle $x^2 + y^2 - 3x + 8y - 4 = 0$ is $(6, -3)$,then what is the other end of the diameter?

Similar Questions

The length of the diameter of the circle which touches the $x-$axis at the point $(1,0)$ and passes through the point $(2,3)$ is:

The square of the length of the tangent from the point $(3, -4)$ to the circle $x^2 + y^2 - 4x - 6y + 3 = 0$ is ....

If the length of the chord $2x + 3y + k = 0$ of the circle $x^2 + y^2 - 2x + 4y - 11 = 0$ is $2\sqrt{3}$,then the sum of all possible values of $k$ is

If one end of a diameter of the circle $x^2 + y^2 + 2x + 4y - 3 = 0$ is $(1, 0)$,then the other end of the diameter is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module