Find the length of the intercept made by the | vedclass

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

Question

(C) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. Substituting the

Vedclass · Accepted Answer

$4 \sqrt{13}$

Vedclass · Answer

(C) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. Substituting the given points $(-4, 3)$ and $(12, -1)$: $(x + 4)(x - 12) + (y - 3)(y + 1) = 0$ $x^2 - 12x + 4x - 48 + y^2 + y - 3y - 3 = 0$ $x^2 + y^2 - 8x - 2y - 51 = 0$. The length of the intercept on the $y$-axis is given by $2 \sqrt{f^2 - c}$,where the equation is $x^2 + y^2 + 2gx + 2fy + c = 0$. Here,$2f = -2 \implies f = -1$ and $c = -51$. Intercept length $= 2 \sqrt{(-1)^2 - (-51)} = 2 \sqrt{1 + 51} = 2 \sqrt{52} = 2 \sqrt{4 	imes 13} = 4 \sqrt{13}$.

Find the length of the intercept made by the circle on the $y$-axis,if the line segment joining the points $(-4, 3)$ and $(12, -1)$ is the diameter of the circle.

Similar Questions

The points $(1,0), (0,1), (0,0)$ and $(2k, 3k), k \neq 0$ are concyclic,if $k$ is

If the lines $x + y = 6$ and $x + 2y = 4$ are diameters of a circle whose diameter is $20$,then the equation of the circle is:

$A$ circle lies in the second quadrant and touches both the axes. If the radius of the circle is $4$,then its equation is

The equation of the circle whose centre lies on the line $x-4y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is

The general second-degree equation $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents a circle if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module