# How do you find the equation of a circle passing through the points (4,3) (-2,-5) and (5,2)?

##### 4 Answers

#### Answer:

#### Explanation:

We can write the equation of the circle in the form:

#ax+by+c=x^2+y^2#

This equation is satisfied by the

So we have:

#{ (4a+3b+c = 25), (-2a-5b+c = 29), (5a+2b+c = 29) :}#

Subtracting the first equation from the second and third to eliminate

#{ (-6a-8b = 4), (a-b=4) :}#

Subtracting

#-14a = -28#

Hence:

#{ (a = 2), (b = -2), (c = 23) :}#

So the equation of our circle can be written:

#2x-2y+23 = x^2+y^2#

Subtracting the left hand side from the right, we find:

#0 = x^2-2x+y^2+2y-23#

#color(white)(0) = x^2-2x+1+y^2+2y+1-25#

#color(white)(0) = (x-1)^2+(y+1)^2-5^2#

So we can write the equation as:

#(x-1)^2+(y+1)^2 = 5^2#

This is (more or less) in the form:

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

where

graph{((x-1)^2+(y+1)^2-5^2)((x-1)^2+(y+1)^2-0.02)((x-4)^2+(y-3)^2-0.01)((x-5)^2+(y-2)^2-0.01)((x+2)^2+(y+5)^2-0.01) = 0 [-12, 12, -7, 4.4]}

#### Answer:

#### Explanation:

Let the center be C( a, b ) and the radius r.

The three points are at the same distance r, from C. So,

#### Answer:

I'd do the algebra similar to the compass and straight-edge construction.

#### Explanation:

The three points define three secant lines. Choose two of the secant lines.

The perpendicular bisectors of the secants intersect at the center.

Once the center is known, the radius is the distance between the center and any one of the given points.

The slope of the secant through

the midpoint of the secant is

So the equation of the perpendicular bisector is

(1)#" "# #y=-x# (details left to reader)

The slope of the secant through

the midpoint of the secant is

So the equation of the perpendicular bisector is

(2)#" "# #y=x-2# (details left to reader)

Finding the point of intersection of lines **(1)** and **(2)** is solving the system

The solution and the center of the circle is

The radius of the circle is the distance between the center

Therefore, the equation of the circle is

#(x-1)^2+(y+1)^2=25#

#### Answer:

#### Explanation:

Here's another solution, specific to the conditions of this particular example...

Let:

#{ (A = (4, 3)), (B = (-2, -5)), (C = (5, 2)) :}#

Calculating the distances between pairs of these points we find:

#d_(AB) = sqrt((-2-4)^2+(-5-3)^2) = sqrt(36+64) = sqrt(100) = 10#

#d_(BC) = sqrt((2+5)^2+(5+2)^2) = sqrt(49+49) = sqrt(98)#

#d_(CA) = sqrt((3-2)^2+(5-4)^2) = sqrt(1+1) = sqrt(2)#

Notice that:

#d_(AB)^2 = 100 = 98 + 2 = d_(BC)^2+d_(CA)^2#

So, by Pythagoras, we can see that

Hence

The midpoint of

#(1/2(4-2), 1/2(3-5)) = (1, -1)#

Hence the equation of the circle can be written:

#(x-1)^2+(y+1)^2 = 5^2#

being in the standard form:

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

with