Question

A circle has a radius of \sqrt{13}√ ​13 ​ ​​ square root of, 13, end square root units and is centered at (-9.3,4.1)(−9.3,4.1)left parenthesis, minus, 9, point, 3, comma, 4, point, 1, right parenthesis. write the equation of this circle.

Answer 1

Answer:  The equation of the circle is

$x^2+y^2+18.6x-8.2y+90.3=0.$

Step-by-step explanation:  We are given to write the equation of the circle with radius √13 units and center at the point (-9.3, 4.1).

We know that

the standard equation of a circle with radius r units and center at the point (h, k) is given by

$(x-h)^2+(y-k)^2=r^2.$

In the given circle,

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

Therefore, the equation of the circle will be

$(x-(-9.3))^2+(y-4.1)^2=(\sqrt{13})^2\\\\\Rightarrow (x+9.3)^2+(y-4.1)^2=13\\\\\Rightarrow x^2+18.6y+86.49+y^2-8.2y+16.81=13\\\\\Rightarrow x^2+y^2+18.6x-8.2y+103.3=13\\\\\Rightarrow x^2+y^2+18.6x-8.2y+90.3=0.$

Thus, the equation of the circle is

$x^2+y^2+18.6x-8.2y+90.3=0.$