Find an equation of the circle that satisfies the given conditions.

Hello sally1,

The equation for a circle is: (x-h)² + (y-k)² = r² 

(h, k) is the center.

So the center of the circle will be halfway between the endpoints of the diameter (the midpoint). Use the Midpoint Formula: [(x₁+x₂)/2 ,  (y₁+y₂)/2]

= [(-1 + 5)/2 , (1 + 9)/2]

= (2, 5)

So the equation so far is: (x-2)² + (y-5)² = r²

Now find the radius, which is the diameter divided in half. 

Use the Distance Formula to find the diameter:

Plug in the endpoints of the diameter: d = sqrt[(5 - -1)² + (9- 1)²]

Simplify.

d = sqrt( 36 + 64)

d = sqrt(100)

d = 10

Half of that = radius = 5

Radius squared = 25

So the final equation: (x-2)² + (y-5)² = 25

$$\\P(-1, 1)\quad and \quad Q(5, 9)\\ \boxed{Center = \frac{1}{2}\left( P+Q\right)} =\frac{1}{2}* \left[ \left(\begin {array} {c} -1 \\1 \end {array} \right) + \left(\begin {array} {c} 5 \\ 9 \end {array} \right) \right]\\ =\frac{1}{2}* \left(\begin {array} {c} -1+5 \\1+9 \end {array} \right)\\ =\frac{1}{2}* \left(\begin {array} {c} 4 \\10 \end {array} \right)\\ =\left(\begin {array} {c} \frac{1}{2}* 4 \\\frac{1}{2}* 10 \end {array} \right)\\ \boxed{Center=\left(\begin {array} {c} 2 \\5 \end {array} \right) =(2, 5) }$$

$$\\\boxed{\vec{radius}= \frac{1}{2}\left( P-Q\right)} =\frac{1}{2}* \left[ \left(\begin {array} {c} -1 \\1 \end {array} \right) - \left(\begin {array} {c} 5 \\ 9 \end {array} \right) \right]\\ =\frac{1}{2}* \left(\begin {array} {c} -1-5 \\1-9 \end {array} \right)\\ =\frac{1}{2}* \left(\begin {array} {c} -6 \\-8 \end {array} \right)\\ =\left(\begin {array} {c} \frac{1}{2}* -6 \\\frac{1}{2}* -8 \end {array} \right)\\ \vec{radius}= \left(\begin {array} {c} -3 \\-4 \end {array} \right)\\ radius=\|\vec{radius}\|=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5\\ \boxed{radius=5}$$

The equation for the circle is: $$(x-2)^2+(y-5)^2=25 \qquad | \quad(x-x_{center})^2+(y-y_{center})^2=r^2$$