Question

Find the equation of the circle that has diameter endpoints of ( 4, 8 ) and (10, 16 )

Answer 1

Answer:

$(x-7)^2+(y-4)^2=25$

Step-by-step explanation:

The circle has diameter endpoints of ( 4, 8 ) and (10, 16 ).

The midpoint of the diameter is the center

$(\frac{4+10}{2},\frac{8+16}{2})$

$(\frac{14}{2},\frac{24}{2})$

$(7,12)$

The radius is the distance from the center to any point on the circumference.

$r=\sqrt{(7-4)^2+(12-8)^2}$

$r=\sqrt{(3)^2+(4)^2}$

$r=\sqrt{9+16}$

$r=\sqrt{25}=5$

The equation of the circle with center (h,k) and radius r is given by:

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

We plug in the values to get:

$(x-7)^2+(y-4)^2=5^2$

$(x-7)^2+(y-4)^2=25$

Answer 2

Answer:

The equation of the circle is (x - 7)² + (y - 12)² = 25

Step-by-step explanation:

* Lets revise the equation of the circle

- The equation of the circle which has center (h , k) is:

 (x - h)² + (y - k)² = r²

- The endpoints of a diameter re (4 , 8) and (10 , 16)

- The center of the circle is the mid-point of the diameter

* Lets revise the rule of the mid point to get the center of the circle

 and revise the rule of the distance to get the length of the diameter

- The mid point (x , y) of segment has endpoints (x1 , y1) and (x2 , y2) is

 x = (x1 + x2)/2  and y = (y1 + y2)/2

- The distance between the two point (x1 , y1) and (x2 , y2) is:

  d = √[(x2 - x1)² + (y2 - y1)²]

* Now lets find the center of the circle

∵ (h , k) is the mid-point of the diameter

∵ (4 , 8) is (x1 , y1) and (10 , 16) is (x2 , y2)

∴ h = (4 + 10)/2 = 14/2 = 7

∴ k = (8 + 16)/2 = 24/2 = 12

∴ The center of the circle is (7 , 12)

∵ The length of the diameter = √[(10 - 4)² + (16 - 8)²

∴ d = √[6² + 8²] = √[36 + 64] = √100 = 10

∵ The radius = 1/2 the diameter

∴ r = 1/2 (10) = 5

* Lets write the equation of the circle

∵ The equation of the circle is (x - h)² + (y - k)² = r²

∵ h = 7 , k = 12 , r = 5

∴ The equation of the circle is (x - 7)² + (y - 12)² = 5²

∴ The equation of the circle is (x - 7)² + (y - 12)² = 25