Respuesta :
Given:
Center of the circle P = (-4,7)
Point on the circle = (-1,9).
To find:
The equation of the circle P.
Solution:
The standard form of a circle is:
[tex](x-h)^2+(y-k)^2=r^2[/tex] ...(i)
Where, (h,k) is the center and r is the radius of the circle.
Center of the circle is at point (-4,7). So, [tex]h=-4,k=7[/tex].
The center passes through the point (-1,9). So, the distance between the point (-1,9) and the center (-4,7) is the radius.
Distance formula:
[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using the distance formula, we get
[tex]r=\sqrt{(-4-(-1))^2+(7-9)^2}[/tex]
[tex]r=\sqrt{(-3)^2+(-2)^2}[/tex]
[tex]r=\sqrt{9+4}[/tex]
[tex]r=\sqrt{13}[/tex]
Substituting [tex]h=-4,k=7[/tex] and [tex]r=\sqrt{13}[/tex] in (i), we get
[tex](x-(-4))^2+(y-7)^2=(\sqrt{13})^2[/tex]
[tex](x+4)^2+(y-7)^2=13[/tex]
Therefore, the equation of the circle P is [tex](x+4)^2+(y-7)^2=13[/tex].
The equation of the circle P is (x +4)² + (y-7)² = 13
What is the standard equation for a circle ?
The standard equation for a circle is
(x - h)² +(y-k)² = r²
here the centre of the circle is (h,k) and r is the radius
It is given that
Circle P is graphed in the coordinate plane with centre (- 4, 7)
The equation can be written as
(x +4)² + (y-7)² = r²
It is also given that
Circle P contains the point (-1,9)
Therefore
( -1 +4)² + (9-7)² = r²
3² + 2² = r²
r² = 9+4 = 13
r = √13
Therefore the equation of the circle P is
(x +4)² + (y-7)² = 13
To know more about standard equation for a circle
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