The equation of the circle of radius 3 that l | vedclass

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

Question

(A) Given,radius $r = 3$. Since the circle lies in the fourth quadrant and touches both coordinate axes ($x=0$ and $y=0$),its center must be at $(h, k

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given,radius $r = 3$. Since the circle lies in the fourth quadrant and touches both coordinate axes ($x=0$ and $y=0$),its center must be at $(h, k) = (3, -3)$. The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the values,we get $(x-3)^2 + (y-(-3))^2 = 3^2$. $(x-3)^2 + (y+3)^2 = 9$. Expanding the terms: $(x^2 - 6x + 9) + (y^2 + 6y + 9) = 9$. $x^2 + y^2 - 6x + 6y + 18 = 9$. $x^2 + y^2 - 6x + 6y + 9 = 0$.

The equation of the circle of radius $3$ that lies in the fourth quadrant and touches the lines $x=0$ and $y=0$ is

Similar Questions

The polar equation of the circle with centre $\left(2, \frac{\pi}{2}\right)$ and radius $3$ units is :

Find the centre and radius of the circle $x^{2}+y^{2}-8x+10y-12=0$.

The parametric equations of the curve $x^2+y^2+ax+by=0$ are

The equation of a circle with centre $(-4, 3)$ and touching the circle $x^2 + y^2 = 1$ is

The points of contact of the circle $x^2 + y^2 + 2x + 2y + 1 = 0$ with the coordinate axes are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module