Question

the equation of the circle with centre at (-3,-2)and radius 6

Answer 1

Answer:

[(x+3)^2 + (y+2)^2 = 36] OR. [x^2 + y^2 +6x + 4y = 23]

Explanation:

given,

centre (-3,-2),radius=6

.: Let other point be (x,y).

Distance formula = √[(x2-x1)^2 + (y2-y1)^2].

6 = √[(x+3)^2 + (√y+2)^2]

.: Squaring both sides,

36 = (x+3)^2 + (y+2)^2 is the equation.

Or

36= x^2 +9+6x + y^2 +4+4y

36=x^2 + y^2 +6x + 4y +13

36 - 13 =x^2 + y^2 +6x + 4y

.: x^2 + y^2 +6x + 4y = 23 is the equation.

Answer 2

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Given that:

Radius, r = 4, and center (h, k) = (-2, 3).

We know that the equation of a circle with centre (h, k) and radius r is given as

$\sf (x – h)^2 + (y – k)^2 = r^2 ….(1)$

Now, substitute the radius and center values in (1), we get

Therefore, the equation of the circle is

$\sf (x + 2)^2+ (y – 3)^2 = (4)^2$

$\sf x^2+ 4x + 4 + y^2 – 6y + 9 = 16$

Now, simplify the above equation, we get:

$\sf x^2 + y^2+ 4x – 6y – 3 = 0$

Thus, the equation of a circle with center (-2, 3) and radius 4 is $\sf x^2 + y^2+ 4x – 6y – 3 = 0$

Hope it's Helpful.....:)