Find the equation of the circle. Centered at (3, -2) with radius … - Vidyadip

MCQ

Find the equation of the circle. Centered at $(3, -2)$ with radius $4:$

A
$ x^2+y^2+6 x-4 y=3 $
✓
$ x^2+y^2-6 x+4 y=3 $
C
$ x^2+y^2-3 x+2 y=-3 $
D
$ x^2+y^2+3 x-2 y=-3 $

✓

Answer

Correct option: B.

$ x^2+y^2-6 x+4 y=3 $

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Conic Sections→Conic Sections — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

The equation of the circle $x^2+ y^2+ 2gx + 2fy + c = 0$ will represent a real circle if:

The AM,HMandGMbetween two numbers are $\frac{144}{15},15$ and 12, but not necessarily in thisorder. Then, HM, GM and AM respectively are:

Logarithm of $32\root 5 \of 4 $ to the base $2\sqrt 2 $ is

Find the sum of squares of first $n$ terms.

The general solution of $sin\, x + sin \,5x = sin\, 2x + sin \,4x$ is :

Let $P$ be a variable point on the parabola $y=4 x^{2}+1$. Then the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y=x$ is:

Range of $f(x) = \frac{{{x^2} + 34x - 71}}{{{x^2} + 2x - 7}}$ is

If $\alpha$ and $\beta$ are the distinct roots of the equation $x^{2}+(3)^{1 / 4} x+3^{1 / 2}=0$, then the value of $\alpha^{96}\left(\alpha^{12}-\right.1) +\beta^{96}\left(\beta^{12}-1\right)$ is equal to:

If $\sin (A + B) =1 $ and $\cos (A - B) = \frac{{\sqrt 3 }}{2},$ then the smallest positive values of $A$ and $ B$ are

A three digit number is formed by using numbers $1, 2, 3$ and $4$. The probability that the number is divisible by $3$, is