How do you find the equation of a circle with center (7,k), radius 5 and with the point (4,3) on the circle.?

1 Answer
Mar 1, 2016

Answer:

#{:((x-7)^2+(y+1)^2=25),((x-7)^2+(y-7)^2=25):}#

Explanation:

The general form of the equation of a circle with a center at #(h,k)# and radius #r# is

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

Since we have a center of #(7,k)# and #r=5#, we get the equation of

#(x-7)^2+(y-k)^2=25#

Now, to determine #k#, we can use the fact that #(4,3)# is a point on the circle:

#(4-7)^2+(3-k)^2=25#

#9+(3-k)^2=25#

#(3-k)^2=16#

Take the square root of both sides. Note that taking the square root will leave a positive and negative version, so we will have two different equations that fit these criteria.

#3-k=+-4#

When #3-k=4#, we see that #k=-1#.

When #3-k=-4#, we see that #k=7#.

Hence the two possible equations for the circle are

#{:((x-7)^2+(y+1)^2=25),((x-7)^2+(y-7)^2=25):}#

graph{((x-7)^2+(y+1)^2-25)((x-7)^2+(y-7)^2-25)((x-4)^2+(y-3)^2-.1)=0 [-18.27, 27.35, -8.16, 14.65]}

The two large circles are the two possible equations. The smaller circle is the point #(3,4)#, which lies on both circles.