A circle has a center that falls on the line #y = 1/4x +2 # and passes through # ( 3 ,7 )# and #(7 ,1 )#. What is the equation of the circle?

Question

A circle has a center that falls on the line #y = 1/4x +2 # and passes through #      (    3  ,7  )# and #(7   ,1  )#.  What is the equation of the circle?

Allison Allen · Accepted Answer

#(x - 16/5)^2+(y-14/5)^2 = (sqrt442/5)^2#

The Cartesian form for the equation of a circle is:

#(x - h)^2+(y-k)^2 = r^2" [1]"#

where #(x,y)# is any point on the circle, #(h,k)# is the center, and #r# is the radius.

We can use the two points, #(3,7)# and #(7,1)# and equation [1] to write two equations:

#(3 - h)^2+(7-k)^2 = r^2" [2]"# #(7 - h)^2+(1-k)^2 = r^2" [3]"#

Expand the squares:

#9 - 6h+h^2+49-14k+k^2 = r^2" [2.1]"# #49 - 14h+h^2+1-2k+k^2 = r^2" [3.1]"#

Subtract equation [3.1] from equation [2.1]:

#8h-12k+8=0" [4]"#

Evaluate the given line at the point #(h,k)#:

#k = 1/4h+2" [5]"#

Substitute equation [5] into equation [4]:

#8h-12(1/4h+2)+8=0#

#8h-3h-24+8=0#

#5h = 16#

#h = 16/5#

Use equation [5] to solve for k:

#k = 1/4(16/5)+2#

#k = 14/5#

Use equation [2] to find the value of r:

#(3 - 16/5)^2+(7-14/5)^2 = r^2#

#(-1/5)^2+(21/5)^2 = r^2#

#1/25+441/25 = r^2#

#r^2= 442/25#

#r = sqrt442/5#

Substitute the know values into equation [1]:

#(x - 16/5)^2+(y-14/5)^2 = (sqrt442/5)^2#

Aurora Ayala · Answer

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

Given that the center falls on the line $ y = \frac{1}{4}x + 2 $, we can set up equations using the coordinates of the points (3, 7) and (7, 1) through which the circle passes.

Let's start by finding the midpoint of the line segment formed by the points (3, 7) and (7, 1). The midpoint formula is:

$$
\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
$$

Using this formula, we find the midpoint to be $\left( \frac{3 + 7}{2}, \frac{7 + 1}{2} \right) = (5, 4)$.

Since the center of the circle falls on the line $ y = \frac{1}{4}x + 2 $, the x-coordinate of the center is 5.

Now, we substitute this x-coordinate into the equation of the line to find the y-coordinate:

$$
y = \frac{1}{4} \times 5 + 2 = \frac{5}{4} + 2 = \frac{5}{4} + \frac{8}{4} = \frac{13}{4}
$$

So, the center of the circle is (5, 13/4).

Next, we find the radius of the circle. The radius is the distance from the center to any point on the circle. We can use the distance formula:

$$
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

Let's choose one of the given points, say (3, 7), and calculate the distance from the center:

$$
d = \sqrt{(3 - 5)^2 + (7 - \frac{13}{4})^2} = \sqrt{(-2)^2 + \left(7 - \frac{13}{4}\right)^2} = \sqrt{4 + \left(\frac{28}{4} - \frac{13}{4}\right)^2} = \sqrt{4 + \left(\frac{15}{4}\right)^2} = \sqrt{4 + \frac{225}{16}} = \sqrt{\frac{641}{16}}
$$

So, the radius is $\sqrt{\frac{641}{16}}$.

Therefore, the equation of the circle is:

$$
(x - 5)^2 + \left(y - \frac{13}{4}\right)^2 = \left(\sqrt{\frac{641}{16}}\right)^2
$$