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A Diameter of a circle has endpoints P(-10, -2) and Q(4, 6)

a. Find the center of the circle.

b. find the radius. If your answer is not an integer, express in in radical form.

c. write an equation for the circle
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Respuesta :

Answer:

see explanation

Step-by-step explanation:

(a)

the centre of the circle is at the midpoint of the endpoints.

using the midpoint formula

centre = ( [tex]\frac{-10+4}{2}[/tex] , [tex]\frac{-2+6}{2}[/tex] ) = ([tex]\frac{-6}{2}[/tex] , [tex]\frac{4}{2}[/tex] ) = (- 3, 2 )

(b)

the radius r is the distance from the centre to either of the endpoints

using the distance formula to find r

r = [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }[/tex]

with (x₁, y₁ ) = (- 3, 2 ) and (x₂, y₂ ) = Q (4, 6 )

r = [tex]\sqrt{(4-(-3))^2+(6-2)^2}[/tex]

  = [tex]\sqrt{(4+3)^2+4^2}[/tex]

  = [tex]\sqrt{7^2+16}[/tex]

  = [tex]\sqrt{49+16}[/tex]

   = [tex]\sqrt{65}[/tex]

(c)

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r the radius , then

(x - (- 3) )² + (y - 2)² = ([tex]\sqrt{65}[/tex] )² , that is

(x + 3)² + (y - 2)² = 65

devishri1977

Answer:

Step-by-step explanation:

a) Use midpoint formula to find the center.

[tex]\sf \boxed{Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)}[/tex]

P(-10 ,-2) ⇒x₁ = -10  & y₁ = -2

Q(4,6) ⇒ x₂=4   & y₂ = 6

[tex]\sf midpoint = \left(\dfrac{-10+4}{2},\dfrac{-2+6}{2}\right)[/tex]

            [tex]=\left(\dfrac{-6}{2},\dfrac{4}{2}\right)\\\\=\left(-3,2\right)[/tex]

b) To find the radius, use the distance formula

[tex]\boxed{distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}[/tex]

Q(4,6)  & O(-3,2)

[tex]r=\sqrt{(-3-4)^2+(2-6)^{2}}\\\\=\sqrt{(-7)^{2}+(-4)^2}\\\\=\sqrt{49+16}\\\\= \sqrt{65}[/tex]

radius = √65

c) Equation of circle : (x - h)² +(y -k)² = r²

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

(x -[-3])² + (y - 2)²= (√65)²

(x + 3)² + (y -2)² = 65

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