Respuesta :

apologiabiology
(x-h)²+(y-k)=r²
radius is r and center is (h,k)
given center at (5,-3)
(x-5)²+(y+3)²=r²
a point is (2,5)
input to find r

(2-5)²+(5+3)²=r²
(-3)²+(8)²=r²
9+64=r²
73=r²

de equation is

(x-5)²+(y+3)²=73

The equation of the circle is [tex]\boxed{{{\left( {x - 5} \right)}^2} + {{\left( {y + 3} \right)}^2} = 73}.[/tex]

Further explanation:

The standard equation of the circle with center [tex]\left( {h,k} \right)[/tex] and radius r can be expressed as,

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

Given:

A circle with center at [tex]\left( { - 3,2} \right).[/tex]

The passing through point is [tex]\left( {1,5} \right).[/tex]

Explanation:

The center is at [tex]\left( {5,-3} \right).[/tex]

The passing through point is [tex]\left( {2,5} \right).[/tex] Therefore, the point satisfies the equation of circle.

Substitute 1 for [tex]x,[/tex] 5 for [tex]y[/tex], [tex]-3[/tex] for [tex]h[/tex] and 2 for [tex]k[/tex] in general equation of the circle to obtain the radius of circle.

[tex]\begin{aligned}{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} &= {r^2}\\{\left( {2 - 5} \right)^2} + {\left( {5 + 3} \right)^2} &= {r^2}\\{\left({ - 3} \right)^2} + {\left( 8 \right)^2} &= {r^2}\\9 + 64 &= {r^2}\\73 &= {r^2}\\\sqrt {73} &= r\\\end{aligned}[/tex]

The equation of the circle is [tex]\boxed{{{\left( {x - 5} \right)}^2} + {{\left( {y + 3} \right)}^2} = 73}.[/tex]

Learn more:

1. Learn more about equation of circle brainly.com/question/1506955.

2. Learn more about domain of the function https://brainly.com/question/3852778.

3. Learn more about coplanar https://brainly.com/question/4165000.

Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Circle

Keywords: Circle, centered point (5,-3), passes point (2,5), standard form of the circle, equation of the circle, center, diameter of circle, radius of the circle,center-radius form, general equation of circle, tangent, area of circle.

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