A circle has a center that falls on the line #y = 11/7x +8 # and passes through # ( 9 ,1 )# and #(8 ,4 )#. What is the equation of the circle?

1 Answer
Mar 26, 2016

#(x+175/26)^2+(y+67/26)^2=87965/338#

Explanation:

Finding the center of the circle:

We can say that the center of the circle lies on the point #(x,y)=(x,11/7x+8)#.

Since we know two points on the circle, we know that the distances from the center to each of the points will be the same.

Use the distance formula from the center to each of these points and set them equal to one another:

#" "sqrt((x-9)^2+((11/7x+8)-1)^2)=sqrt((x-8)^2+((11/7x+8)-4)^2)#

Square both sides and simplify inside the square roots.

#" "(x-9)^2+(11/7x+7)^2=(x-8)^2+(11/7x+4)^2#

Expand.

#" "(x^2-18x+81)+(121/49x^2+22x+49)=(x^2-16x+64)+(121/49x^2+88/7x+16)#

The #x^2# and #121/49x^2# terms present on both sides will cancel. Now, combine like terms on each side.

#" "4x+130=-112/7x+88/7x+80#

#" "4x+130=-24/7x+80" "=>" "28/7x+24/7x=-50#

#" "52/7x=-50" "=>" "52x=-350#

#" "x=-350/52=-175/26#

Plug this value of #x#, which is the #x# coordinate of the circle's center, into the equation of the line to find the #y# coordinate.

#" "y=11/7x+8" "=>" "y=11/7(-175/26)+8#

#" "y=11(-25/26)+208/26" "y=-275/26+208/26#

#" "y=-67/26#

This gives us

#color(blue)(barul|" center "=(-175/26,-67/26)" "|#

Now, we need to determine our next step. Our final goal is to determine the equation of the circle. The standard form of the equation of the circle is

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

Where the circle's center is #(h,k)# and its radius is #r#.

Since we know the circle's center, we know that

#" "color(blue)(barul|{:(h=-175/26),(k=-67/26):}|#

Thus, all we need to do is to find the radius of the circle. (Really, we need to find #r^2#, since that's how it's represented in the equation of the circle.)

Finding the circle's radius:

The radius is the distance from the center to any point on the circle. We can apply the distance formula from #(-175/26,-67/26)# to #(9,1)#.

#" "r=sqrt((9-(-175/26))^2+(1-(-67/26))^2)#

Squaring both sides, so we have #r^2# instead of #r#, we obtain

#" "r^2=(9+175/26)^2+(1+67/26)^2#

#" "r^2=(234/26+175/26)^2+(26/26+67/26)^2#

#" "r^2=(409/26)^2+(93/26)^2#

#" "color(blue)(barul|r^2color(black)(=167281/676+8649/676=175390/676)=87965/338|)#

Constructing the equation of the circle:

Using the equation

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

We have

#" "(x-(-175/26))^2+(y-(-67/26))^2=87965/338#

Slightly simplified:

#" "color(blue)(barul|(x+175/26)^2+(y+67/26)^2=87965/338|)#

Graphical check:

Graphed are the circle, the line #y=11/7x+8#, and the points #(-175/26,-67/26)#, #(9,1)# and #(8,4)#:

graph{((x+175/26)^2+(y+67/26)^2-87965/338)(y-11/7x-8)((x-9)^2+(y-1)^2-.2)((x-8)^2+(y-4)^2-.2)((x+175/26)^2+(y+67/26)^2-.2)=0 [-41.56, 31.5, -21.63, 14.92]}