a circle has a radius of square root of 13 units and is centered at (-9.3,4.1) Write the equation of this circle. Can anyone please help me with this problem

Respuesta :

The equation of this circle is
(x+9.3)² + (y-4.1)² = 13

The format for the equation of a circle is
(x-h)² + (y-k)² = r², where (h, k) is the center of the circle and r is the radius.

The equation of a circle is [tex]\rm x^2+y^2+18.6x-8.2y+90.3=0[/tex].

What is the equation of a circle?

The standard equation of a circle is given by:

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

Where (h,k) is the coordinates of the center of the circle and r is the radius.

A circle has a radius of the square root of 13 units and is centered at (-9.3,4.1).

In the given circle, radius, r = √13 units  and center, (h, k) = (-9.3, 4.1).

The equation of this circle is given;

[tex]\rm (x-(-9.3))^2 + (y-4.1)^2 = \sqrt{13}^2\\\\(x+9.3)^2+(y-4.1)=13\\\\x^2+18.6x+86.49+y^2-8.2y+16.81=13\\\\x^2+y^2+18.6x-8.2y+103.0-12=0\\\\x^2+y^2+18.6x-8.2y+90.3=0[/tex]

Hence, the required equation of a circle is [tex]\rm x^2+y^2+18.6x-8.2y+90.3=0[/tex].

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