Question

Find the equation of circle which passes through (0, 0), (2, 0), (0, 3).​

Answer 1

Answer:

Step-by-step explanation:

Given Points ( 0 , 0 ) , ( 0 , 3 ) and ( 2 , 0 )

Let the equation of circle passing through given point is  

( x - h )² + ( y - k )² = r²

Substituting  ( 0 , 0 )

( 0 - h )² + ( 0 - k )² = r²

h² + k² = r²  (1)

Substituting ( 0 , 3 )

( 0 - h )² + ( 3 - k )² = r²

h² + ( 3 - k )² = r² (2)

Substituting ( 2 , 0 )

( 2 - h )² + ( 0 - k )² = r²

( 2 - h )² + k² = r² (3)

Subtract (1) from (2) -

( 3 - k )² - k² = 0

9 + k² - 6k - k² = 0

-6k + 9 = 0

k = 9/6 = 3/2

Subtracting (1) from (3)

( 2 - h )² - h² = 0

4 + h² - 4h - h² = 0

4h = 4

h = 1

Substituting value of h and k in eqn (1) -

r² = 1²+(3/2)²

r² = 1+9/4

r² = 13/4

r = √13/4

Thus the equation of circle is = (x-1)²+(y-3/2)² = 13/4