Find the equation of the circle with center ( | vedclass

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

Question

(A) The center of the circle is $(h, k) = (2, 1)$.Since the circle touches the $X$-axis,the radius $r$ is equal to the absolute value of the $y$-coord

Vedclass · Accepted Answer

$x^2 + y^2 - 4x - 2y + 4 = 0$

Vedclass · Answer

(A) The center of the circle is $(h, k) = (2, 1)$.Since the circle touches the $X$-axis,the radius $r$ is equal to the absolute value of the $y$-coordinate of the center.Thus,$r = |1| = 1$.The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - 2)^2 + (y - 1)^2 = 1^2$.Expanding this,we have $(x^2 - 4x + 4) + (y^2 - 2y + 1) = 1$.Simplifying,$x^2 + y^2 - 4x - 2y + 4 = 0$.

Find the equation of the circle with center $(2, 1)$ and touching the $X$-axis.

Similar Questions

The equation of the circle passing through $(4, 5)$ and having the centre at $(2, 2)$ is:

The equation of the circle touching $x = 0$,$y = 0$ and $x = 4$ is

The $x$-coordinates of two points $A$ and $B$ are the roots of the equation $x^{2} + 2ax - b^{2} = 0$,and their $y$-coordinates are the roots of the equation $y^{2} + 2py - q^{2} = 0$. Find the radius of the circle with $AB$ as its diameter.

$A$ circle passing through $(0,0)$, $(2,6)$, and $(6,2)$ cuts the $x$-axis at the point $P \neq (0,0)$. Then, the length of $OP$, where $O$ is the origin, is

The circle passing through $(1, -2)$ and touching the $x$-axis at $(3, 0)$ also passes through the point

Vedclass Test Series

Exam Paper Generator

Online Exam Module