Find the equation of the circle with centre (–2, 3) and radius 4

Open in App

Solution

The equation of a circle with centre (h, k) and radius r is given as

(x – h)² + (y – k)² = r²

It is given that centre (h, k) = (–2, 3) and radius (r) = 4.

Therefore, the equation of the circle is

(x + 2)² + (y – 3)² = (4)²

x² + 4x + 4 + y² – 6y + 9 = 16

x² + y² + 4x – 6y – 3 = 0

Suggest Corrections

Similar questions

Find the equation of the circle with
centre (-2,3) and radius 4

Q. Find the equation of the circle with radius

5

whose centre lies on x-axis and passes through the point

(2, 3)

Q. Find the equation of the circle with:
(i) Centre (−2, 3) and radius 4.
(ii) Centre (a, b) and radius

\sqrt{a^{2} + b^{2}}

.
(iii) Centre (0, −1) and radius 1.
(iv) Centre (a cos α, a sin α) and radius a.
(v) Centre (a, a) and radius

\sqrt{2}

Find the equation of the circle with:
(i) Centre (-2,3 and radius 4.
(ii) Centre (a, b) and radius $\sqrt{a^{2} + b^{2}}$
(iii) Centre (0,-1) and radius 1.
(iv) Centre $(α c o s α, α s i n α)$ and radius $α$ .
(v) Centre $(α, a)$ and radius $\sqrt{2} a$ .

Q. The axes being inclined at

45^{\circ}

, find the equation to the circle whose centre is the point

(2, 3)

and whose radius is

4

Join BYJU'S Learning Program

Find the equation of the circle with centre (–2, 3) and radius 4

The equation of a circle with centre (h, k) and radius r is given as (x­ – h)2 + (y ­– k)2 = r2 It is given that centre (h, k) = (–2, 3) and radius (r) = 4. Therefore, the equation of the circle is (x + 2)2 + (y – 3)2 = (4)2 x2 + 4x + 4 + y2 – 6y + 9 = 16 x2 + y2 + 4x – 6y – 3 = 0