A circle touches both axes and its center lie | vedclass

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

Question

(A) Since the circle touches both axes and lies in the fourth quadrant with radius $r = 1$,the center of the circle must be $(1, -1)$.The standard equ

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Since the circle touches both axes and lies in the fourth quadrant with radius $r = 1$,the center of the circle must be $(1, -1)$.The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $h = 1$,$k = -1$,and $r = 1$:$(x - 1)^2 + (y - (-1))^2 = 1^2$$(x - 1)^2 + (y + 1)^2 = 1$Expanding the equation:$(x^2 - 2x + 1) + (y^2 + 2y + 1) = 1$$x^2 + y^2 - 2x + 2y + 2 = 1$$x^2 + y^2 - 2x + 2y + 1 = 0$

$A$ circle touches both axes and its center lies in the fourth quadrant. If its radius is $1$,then its equation is:

Similar Questions

The radius of the circle $r = \sqrt{3} \sin \theta + \cos \theta$ is

The radius of a circle which touches the $y$-axis at $(0,3)$ and cuts an intercept of $8$ units with the $x$-axis is

If the lines $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$ lie along diameters of a circle of circumference $10\pi$,then the equation of the circle is

Find the equation of a circle that touches the lines $x = 0$,$y = 0$,and $x = 2c$.

Find the equation of a circle whose diameters are $2x - 3y = 5$ and $3x - 4y = 7$ and whose radius is $8$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module