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?

Question

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?

Maverick Smith · Accepted Answer

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

locating the circle's center:

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

We are aware that the distances between the center and each of the two points on the circle will be equal because we know two of the points.

Set each of these points equal by using the formula for the distance between the center and each of these points:

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

Simplify inside the square roots and square both sides.

#" "(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 provides us with

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

The next thing we must do is figure out what to do next. In the end, we want to figure out the circle's equation, and the standard form of that equation is

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

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

Knowing the center of the circle allows us to 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 radius of the circle:

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|)#

Formulating the circle's equation:

Applying the formula

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

We have

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

Slightly streamlined:

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

Visual inspection:

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

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)^2-87965/338)(x+175/26)^2+(y+67/26)^2=0 [-41.56, 31.5, -21.63, 14.92]}

Sadie Allen · Answer

undefined To find the equation of the circle, we first need to find the center of the circle. Since the center falls on the line $y = \frac{11}{7}x + 8$, we can equate the equation of the line with the general form of a circle's equation, which is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius.

Given that the circle passes through the points $(9, 1)$ and $(8, 4)$, we can find the midpoint of these two points, which will be the center of the circle.

Midpoint formula:
$$h = \frac{x_1 + x_2}{2}$$
$$k = \frac{y_1 + y_2}{2}$$

Substitute the coordinates of the given points:
$$h = \frac{9 + 8}{2} = \frac{17}{2}$$
$$k = \frac{1 + 4}{2} = \frac{5}{2}$$

So, the center of the circle is $(\frac{17}{2}, \frac{5}{2})$.

Now, we need to find the radius of the circle. We can use the distance formula between the center of the circle and one of the given points. Let's use the point $(9, 1)$ for this calculation.

Distance formula:
$$r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Substitute the coordinates:
$$r = \sqrt{(9 - \frac{17}{2})^2 + (1 - \frac{5}{2})^2}$$
$$r = \sqrt{(\frac{1}{2})^2 + (-\frac{3}{2})^2}$$
$$r = \sqrt{\frac{1}{4} + \frac{9}{4}}$$
$$r = \sqrt{\frac{10}{4}} = \frac{\sqrt{10}}{2}$$

Now, we have the center $(\frac{17}{2}, \frac{5}{2})$ and the radius $\frac{\sqrt{10}}{2}$. We can plug these values into the general form of the circle's equation to obtain the final equation:

$$(x - \frac{17}{2})^2 + (y - \frac{5}{2})^2 = (\frac{\sqrt{10}}{2})^2$$

This simplifies to:

$$(x - \frac{17}{2})^2 + (y - \frac{5}{2})^2 = \frac{10}{4}$$

Therefore, the equation of the circle is:

$$(x - \frac{17}{2})^2 + (y - \frac{5}{2})^2 = \frac{5}{2}$$