The points (3,12) (3,2) and (20,-5) lie on a circle.

Question

Find the equation of the circle.? 

 

 

Answer 1

The three points are (3, 12) , (3,2) and (20, -5)

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

Now substitute (3 , 12)

(3 - h)² + (12 - k)² = r²

9 + h² - 6h + 144 + k² - 24k = r²

h² + k² - 6h - 24k +153 = r²    ..................Equation(1)

Now substitute (3 , 2)

(3 - h)² + (2 - k)² = r²

9 + h² - 6h + 4 + k² - 4k = r²

h² + k² - 6h - 4k + 13 = r²       ..................Equation(2)

Now substitute (20 , -5)

(20 - h)² + (-5 - k)² = r²

400 + h² - 40h + 25 + k² +10k = r²

h² + k² - 40h + 10k +425 = r²   ...............Equation(3)

Now Equate Equation (1) and (2)

h² + k² - 6h -24k +153 = h² + k² - 6h - 4k + 13

-24k +153 = - 4k + 13

20k = 140

k = 7

Now Equate Equation (2) and (3)

h² + k² - 6h - 4k + 13 = h² + k² - 40h + 10k +425

- 6h - 4k + 13 = -40h + 10k +425

40h - 10k - 6h - 4k = 425 - 13

34h - 14k = 412

Substitute k = 7

34h - 14*7 = 412

34h = 510

h = 15

Therefore h = 15 and k = 7, substitute in equation(1)

15² + 7² - 6*15 - 24*7 +153 = r²

r² = 169

r = 13.

The Circle Equation is (x - h)² + (y - k)² = r².

(x - 15)² + (y - 7)² = 169

Therefore Equation of Circle is (x - 15)² + (y - 7)² = 169.