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

4 Answers
Feb 11, 2017

Answer:

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

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 #(x, y)# pairs #(4, 3)#, #(-2, -5)# and #(5, 2)#

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 #c#, we get:

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

Subtracting #8# times the second of these equations from the first, we get:

#-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 #(h, k) = (1, -1)# is the centre of the circle and #r=5# the radius:

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]}

Feb 11, 2017

Answer:

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

Explanation:

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

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

#r^2=(a-4)^2+(b-3)^2=(a+2)^2+(b+5)^2=(a-5)^2+(b-2)^2#, giving

#3a+4b+1=0 and b = -a#. Solving,

#a = 1, b=-1 and r = 5#. So, the equation of the circle is

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

Feb 11, 2017

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 #(-2,-5)# and #(5,2)# is
#m = 7/7 = 1#,
the midpoint of the secant is #(3/2,-3/2)# and the slope of the perpendicular is #m_"perp" = -1#

So the equation of the perpendicular bisector is

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

The slope of the secant through #(5,2)# and #(4,3)# is
#m = 1/-1 = -1#,
the midpoint of the secant is #(9/2,5/2)# and the slope of the perpendicular is #m_"perp" = 1#

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

#y=-x#
#y=x-2#.

The solution and the center of the circle is #(1,-1)#

The radius of the circle is the distance between the center #(1,-1)# and any of the three given points.

#r = 5# (details left to the reader)

Therefore, the equation of the circle is

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

Feb 11, 2017

Answer:

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

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 #ABC# is a right angled triangle, with hypotenuse #AB#

Hence #AB# is a diameter of its circumscribing circle, which has centre at the midpoint of #AB# and radius #1/2*d_(AB) = 10/2 = 5#

The midpoint of #AB# is:

#(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 #(h, k) = (1, -1)# the centre and #r=5# the radius.