A circle has a center that falls on the line #y = 7/9x +7 # and passes through # ( 7 ,3 )# and #(5 ,1 )#. What is the equation of the circle?

Question

A circle has a center that falls on the line #y = 7/9x +7 # and passes through #      (    7  ,3  )# and #(5   ,1  )#.  What is the equation of the circle?

Ella Armstrong · Accepted Answer

THe equation of the circle is #(x-11/16)^2+(y-117/16)^2=58.4# Let the center of the circle be #a,b# Then the equation of the circle is #(x-a)^2+(y-b)^2=r^2# Putting the points #(7,3)# and #(5,1)# in the equation of the circle #(7-a)^2+(3-b)^2=r^2# and #(5-a)^2+(1-b)^2=r^2# so #(7-a)^2+(3-b)^2=(5-a)^2+(1-b)^2# Expanding the equations #49-14a+a^2+9-6b+b^2=25-10a+a^2+1-2b+b^2# #49-14a+9-6b=25-10a+1-2b# #58-14a-6b=26-10a-2b# #32=4a+4b# #=># #8=a+b# this is equation 1 From the equation of the line, we have #b=(7a)/9+1# #9b=7a+9# this is equation 2 Solving for a and b , we get #a=11/16# and #b=117/16# Then we calculate #r# #r^2=(7-11/16)^2+(3-117/16)^2# #r^2=(101^2+69^2)/16# #r=7.65# So the equation of the circle is #(x-11/16)^2+(y-117/16)^2=58.4#

Parker Allen · Answer

undefined To find the equation of the circle, we first need to find its center and radius.

The center of the circle lies on the line $y = \frac{7}{9}x + 7$. We can find the slope of this line, which is $ \frac{7}{9} $.

The perpendicular bisector of the line segment joining the two given points will pass through the center of the circle. The midpoint of the segment is $ \left( \frac{7 + 5}{2}, \frac{3 + 1}{2} \right) = (6, 2) $.

The slope of the line passing through the two given points is $ \frac{1 - 3}{5 - 7} = -1 $. The negative reciprocal of this slope is $ 1 $.

Using the point-slope form, the equation of the perpendicular bisector passing through (6, 2) is:

$ y - 2 = 1 \cdot (x - 6) $  
$ y - 2 = x - 6 $  
$ y = x - 4 $

Solving the system of equations for $ y $, we have:

$ \frac{7}{9}x + 7 = x - 4 $  
$ \frac{7}{9}x - x = -4 - 7 $  
$ -\frac{2}{9}x = -11 $  
$ x = \frac{99}{2} $

Substituting this value back into the equation of the line to find $ y $:

$ y = \frac{7}{9} \cdot \frac{99}{2} + 7 $  
$ y = \frac{77}{2} + 7 $  
$ y = \frac{77 + 14}{2} $  
$ y = \frac{91}{2} $

So, the center of the circle is $ \left( \frac{99}{2}, \frac{91}{2} \right) $.

The radius of the circle is the distance from the center to either of the given points. Using the distance formula, the radius is:

$ r = \sqrt{(99/2 - 7)^2 + (91/2 - 3)^2} $  
$ r = \sqrt{(85/2)^2 + (85/2)^2} $  
$ r = \sqrt{2} \cdot \frac{85}{2} $  
$ r = \frac{85 \sqrt{2}}{2} $

Therefore, the equation of the circle is:

$ (x - \frac{99}{2})^2 + (y - \frac{91}{2})^2 = \left( \frac{85 \sqrt{2}}{2} \right)^2 $