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The radius of a circle is 3√2 units and its center is at (-2, 4). Find the equation of the circle.



can you answer this please


Sagot :

Answer:

Equation of the circle: [tex]x^2 + y^2 +4x -8y +2 = 0[/tex]

Step-by-step explanation:

Given:

[tex]r = 3\sqrt{2} \\C(-2, 4)[/tex]

Find:

Equation of the Circle= ?

Solution:

Standard form of Circle:

[tex](x-h)^{2} + (y-k)^{2} = r^{2}[/tex]

The General form of Circle:

[tex]x^2 + y^2 + Dx + Ey + F = 0[/tex]

where D, E and F are constants.

since we are given the points (or coordinates of the circle):

C(h, k) = C(-2, 4)

We can substitute the values of h, k and the radius in the standard form:

[tex][x-(-2)]^{2} + [y- (4)]^{2} = (3\sqrt{2} )^2\\(x+2)^{2} + (y-4)^{2} = 18\\(x^{2} +2x + 2x +4) +(y^2 -4y - 4y +16) = 18\\(x^2 +4x+4) + (y^2 -8y + 16)=18\\[/tex]

Combine like terms:

[tex]x^2 + y^2+4x -8y + 4+16-18 = 0\\\\x^2 +y^2 +4x-8y +2 = 0[/tex]←  (This is the equation of the circle)