The equation of the circle which touches the | vedclass

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

Question

(D) Since the circle touches the $X$-axis at $(1, 0)$,the $x$-coordinate of the center is $1$. Since the circle touches the $Y$-axis at $(0, 1)$,the $

Vedclass · Accepted Answer

$x^2+y^2-2x-2y+1=0$

Vedclass · Answer

(D) Since the circle touches the $X$-axis at $(1, 0)$,the $x$-coordinate of the center is $1$. Since the circle touches the $Y$-axis at $(0, 1)$,the $y$-coordinate of the center is $1$. Thus,the center of the circle is $(1, 1)$ and the radius is $r = 1$. The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the values,we get $(x-1)^2 + (y-1)^2 = 1^2$. Expanding this,we have $(x^2 - 2x + 1) + (y^2 - 2y + 1) = 1$. Simplifying,we get $x^2 + y^2 - 2x - 2y + 1 = 0$.

The equation of the circle which touches the $X$-axis and $Y$-axis at the points $(1, 0)$ and $(0, 1)$ respectively is

Similar Questions

The parametric equations of the circle $x^2+y^2-6x-2y+9=0$ are

The equation of the circle concentric with the circle $x^2 + y^2 + 8x + 10y - 7 = 0$ and passing through the centre of the circle $x^2 + y^2 - 4x - 6y = 0$ 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.

The equation of the circle which touches the $x$-axis at $(3, 0)$ and passes through $(1, 4)$ is given by

The number of circles touching the lines $x = 0$,$y = a$ and $y = b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module