Write an equation of the circle whose DIAMETER is (5,4) and (-3,2).?

Diameter distance will be given by

d = √ [ (5 - (-3))² + (4 - 2)²]

d = √68

=> radius ( r ) = √68/2 = √17

Now centre of the circle is the mid-point of the given extremes of the diameter and is given by

Center (h,k) = (5 - 3)/2, (4 + 2)/2 = (1.3)

Equation of the circle with centre (1,3) and radius √17 is,

(x - 1)² + (y - 3)² = √17²

x² + 1 - 2x + y² + 9 - 6y = 17

x² + y² - 2x - 6y - 7 = 0

Use the distance formula to find the diameter:

d = SQR[(x2 - x1)^2 + (y2 - y1)^2] = SQR[(-3 - 5)^2 + (2 - 4)^2]

= SQR(64 + 4) = SQR(68) = 2SQR(17) ~ 8.246

Radius = d/2 = 4.123

Use the midpoint formula to find the center of the circle:

[(x1 + x2)/2 , (y1 + y2)/2] =

[(5 + (-3))/2, (4 + 2)/2] = (1, 3)

A circle with radius = r and center (h, k):

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

(x - 1)^2 - (y - 3)^2 = 17

- .--

C (1 , 3)

d² = 8² + 2² = 68

d = √68

r = √68 / 2

r² = 68/4 = 17

(x - 1)² + (y - 3)² = 17

x1 = 5

y1 = 4

x2 = - 3

y2 = 2

Center = C (x, y):

C [(x1 + x2) / 2, (y1 + y2) / 2]

C [(5 - 3) / 2, (4 + 2) / 2]

C (2/2, 6/2)

C (1, 3)

h = 1

k = 3

Diameter = d:

d² = (x2 - x1)² + (y2 - y1)²:

d² = (- 3 - 5)² + (2 - 4)²

d² = (- 8)² + (- 2)²

d² = 64 + 4

d² = 68

d = √68

d = 8.246

Radius = r:

r = 1/2 d:

r = 1/2 (8.246)

r = 4.123

Equation:

(x - h)² + (y - k)² = r²:

(x - 1)² + (y - 3)² = (4.123)²

(x - 1)² + (y - 3)² = 17

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

The centre is ( (5-3)/2, (4+2)/2 ) = (1, 3)

and the diameter is sqrt( (-3-5)² + (2-4)²) = sqrt(64 + 4) = sqrt(68)

so the radius is ½sqrt(68) = sqrt(17)

Can you do it now?

Step 1: Prove the Riemann Hypothesis.

Thus, the problem trivializes from this step.