Find the equation of the circle with centre $(-3, 2)$ and radius $4$.
- A$(x+3)^{2} + (y-2)^{2} = 16$
- B$(x-3)^{2} + (y+2)^{2} = 16$
- C$(x+3)^{2} + (y+2)^{2} = 16$
- D$(x-3)^{2} + (y-2)^{2} = 16$
Explore More
Similar Questions
The equation of the circle whose centre lies on the line $x-4y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is
The equation of the circle,concentric with the circle $x^2+y^2-6x-4y-12=0$ and touching the $X$-axis is
The equation of the circle having its centre on the line $2x + y + 3 = 0$ and having the lines $3x + 4y - 18 = 0$ and $3x + 4y + 2 = 0$ as tangents is:
The equation of the circle whose end points of a diameter are the centres of the circles $x^{2}+y^{2}+2x-4y+1=0$ and $x^{2}+y^{2}-8x+6y+17=0$ is
If the lines $x+2y-5=0$ and $2x-3y+4=0$ lie along diameters of a circle of area $9\pi$,then the equation of the circle is
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo