Type the correct answer in each box.
A circle is centered at the point (5, -4) and passes through the point (-3, 2).
The equation of this circle is (x + ___)^2 + (y + ___)^2 = ___.

Answer :

luisejr77

Answer: [tex](x-5)^2+(y+4)^2=10^2[/tex]

Step-by-step explanation:

The standard of equation of a circumference has the form:

[tex](x-a)^2+(y-b)^2=r^2[/tex]

Where the point (a,b) is the center of the circumference and r is the radius.

You know the center of the circumference: (5,-4).

Substitute this point into the equation of the circumference:

a=5 and b=-4

Then:

[tex](x-5)^2+(y+4)^2=r^2[/tex]

Now you need to find the radius. Substitute the point (-3,2) into the circumference and solve for r:

[tex](-3-5)^2+(2+4)^2=r^2\\(-8)^2+(6)^2=r^2\\64+36=r^2\\r=\sqrt{100}\\r=10[/tex]

The equation of this circumference is:

 [tex](x-5)^2+(y+4)^2=10^2[/tex]

Answer:

[tex](x-5)^2+(y+4)^2=100[/tex]

Or type

(x + _-5__)^2 + (y + _4__)^2 = _100__.

Step-by-step explanation:

First find the radius of the circle using the distance formula;

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

Substitute the points to get;

[tex]r=\sqrt{(5--3)^2+(-4-2)^2}[/tex]

[tex]r=\sqrt{8^2+(-6)^2}[/tex]

[tex]r=\sqrt{64+36}[/tex]

[tex]r=\sqrt{100}[/tex]

r=10 units

We now substitute the center (h,k)=(5,-4) and r=10 into the standard equation of the circle;

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

[tex](x-5)^2+(y--4)^2=10^2[/tex]

[tex](x-5)^2+(y+4)^2=100[/tex]

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