A circle has a center that falls on the line #y = 7/6x +1 # and passes through #(9 ,4 )# and #(8 ,5 )#. What is the equation of the circle?

Question

A circle has a center that falls on the line #y = 7/6x +1 # and passes through #(9  ,4  )# and #(8   ,5  )#.  What is the equation of the circle?

Maverick Smith · Accepted Answer

#(x+31)^2+(y+35)^2=3121# The straight #y = 7/6x+1# can be represented as #(p-p_0).vec v =0# where #p = (x,y), p_0 = (-1,0)# and #vec v = (7,-6)# and also can be represented in the parametric form as: #S_1-> p = p_0+lambda* vec v^T# where #vec v^T# is the vector with components #(6,7)# which is orthogonal to #vec v#. The center point to the circle is at the intersection of #S_1# and #S_2# where #S_2# is the mediatrix to the segment #bar(p_1p_2)#. #S_2# is written as: #S_2->p = p_{12}+mu*vec v_{12}^T# where #p_{12} = (p_1+p_2)/2,vec v_{12} = p_1-p_2 = (1,-1)# and #vec v_{12}^T = {-1,-1}#. Solving #p_0+lambda_c* vec v^T = p_{12}+mu_c*vec v_{12}^T# for #mu_c, lambda_c# we obtain #mu_c= -5, \lambda_c = 79/2# Now, putting all together, #p_c = p_0+lambda_c* vec v^T =(-31, -35) # #r = norm(p_c-p_1)=sqrt[3121]#

Sadie Allen · Answer

undefined The equation of the circle is $ (x - h)^2 + (y - k)^2 = r^2 $, where $ (h, k) $ is the center of the circle and $ r $ is the radius.

First, find the center of the circle using the given line equation:
$$ y = \frac{7}{6}x + 1 $$
$$ \frac{7}{6}x = y - 1 $$
$$ x = \frac{6}{7}(y - 1) $$
$$ h = \frac{6}{7}(k - 1) $$

Then, find the radius ($ r $) using the distance formula between the center and one of the given points:
$$ r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} $$

Substitute the coordinates of one of the given points into the distance formula to find $ r $.

Finally, substitute the values of $ h $, $ k $, and $ r $ into the equation $ (x - h)^2 + (y - k)^2 = r^2 $ to get the equation of the circle.