The equation of the circle concentric with th | vedclass

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

Question

(B) Step $1$: Find the center of the given circle $x^2 + y^2 - 6x + 7 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$, we get $g = -3$ and $f = 0

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Step $1$: Find the center of the given circle $x^2 + y^2 - 6x + 7 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$, we get $g = -3$ and $f = 0$. The center is $(-g, -f) = (3, 0)$.Step $2$: Since the required circle is concentric, its center is also $(3, 0)$. Let its equation be $x^2 + y^2 - 6x + k = 0$.Step $3$: The radius $r$ of this circle is the perpendicular distance from the center $(3, 0)$ to the line $x + y + 3 = 0$. Using the formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$, we get $r = \frac{|1(3) + 1(0) + 3|}{\sqrt{1^2 + 1^2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.Step $4$: The radius squared is $r^2 = (3\sqrt{2})^2 = 18$. The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Substituting $(3, 0)$ and $r^2 = 18$, we get $(x-3)^2 + (y-0)^2 = 18$, which simplifies to $x^2 - 6x + 9 + y^2 = 18$, or $x^2 + y^2 - 6x - 9 = 0$.

The equation of the circle concentric with the circle $x^2 + y^2 - 6x + 7 = 0$ and which touches the line $x + y + 3 = 0$ is:

Similar Questions

If $(a, 1/a)$,$(b, 1/b)$,$(c, 1/c)$,and $(d, 1/d)$ are four distinct points on a circle of radius $4$ units,then $abcd$ is equal to

The equation of the circle with centre $(2, -3)$ and touching the $X$-axis is

The circle passing through the point $(-1, 0)$ and touching the $y$-axis at $(0, 2)$ also passes through the point

The radius of the circle $(x - 1)(x - 3) + (y - 2)(y - 4) = 0$ is

The equation of the circle having centre $(1, -2)$ and passing through the point of intersection of lines $3x + y = 14$ and $2x + 5y = 18$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module