The equation of the circle which has its cent | vedclass

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

Question

(A) The radius $r$ of the circle is the perpendicular distance from the center $(3, 4)$ to the line $5x + 12y - 11 = 0$. $r = \left| \frac{5(3) + 12(4

Vedclass · Accepted Answer

$x^2 + y^2 - 6x - 8y + 9 = 0$

Vedclass · Answer

(A) The radius $r$ of the circle is the perpendicular distance from the center $(3, 4)$ to the line $5x + 12y - 11 = 0$. $r = \left| \frac{5(3) + 12(4) - 11}{\sqrt{5^2 + 12^2}} \right| = \left| \frac{15 + 48 - 11}{\sqrt{25 + 144}} \right| = \frac{52}{13} = 4$. The equation of the circle with center $(h, k) = (3, 4)$ and radius $r = 4$ is given by $(x - h)^2 + (y - k)^2 = r^2$. $(x - 3)^2 + (y - 4)^2 = 4^2$ $x^2 - 6x + 9 + y^2 - 8y + 16 = 16$ $x^2 + y^2 - 6x - 8y + 9 = 0$.

The equation of the circle which has its centre at the point $(3, 4)$ and touches the line $5x + 12y - 11 = 0$ is

Similar Questions

The equation of the concentric circle,with the circle $C_1$ having equation $x^2+y^2-6x-4y-12=0$ and having double the area compared to the area of $C_1$,is

$A$ circle passes through the point $(1, -2)$ and touches the $x$-axis at $(3, 0)$. Through which of the following points does it also pass?

The radius of the circle $r=12 \cos \theta+5 \sin \theta$ is

If four distinct points $(2k, 3k)$, $(2,0)$, $(0,3)$, and $(0,0)$ lie on a circle, then:

The equation of a circle passing through the origin and making an $x$-intercept of $3$ and a $y$-intercept of $-5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module