Question

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

Answer 1

Answer:

Equation of Circle is $(x-1)^2+(y-\frac{3}{2})^2=\frac{13}{4}$

Step-by-step explanation:

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

To find: Equation Circle

Let the equation of circle passing through given point is  

( x - h )² + ( y - k )² = r²

Now we find value of h , k , r by point given point and solving the.

First put ( 0 , 0 )

we get ( 0 - h )² + ( 0 - k )² = r²

h² + k² = r² ........... (1)

Second put ( 0 , 3 )

we get,  ( 0 - h )² + ( 3 - k )² = r²

h² + ( 3 - k )² = r² ........(2)

Third put ( 2 , 0 )

we get ( 2 - h )² + ( 0 - k )² = r²

( 2 - h )² + k² = r² ..............(3)

Subtract (1) from (2)

we get,

( 3 - k )² - k² = 0

9 + k² - 6k - k² = 0

-6k + 9 = 0

$k=\frac{9}{6}$

$k=\frac{3}{2}$

Now, Subtract (1) from (3)

( 2 - h )² - h² = 0

4 + h² - 4h - h² = 0

4h = 4

h = 1

put value of h & k in eqn (1) we get

$r^2=1^2+(\frac{3}{2})^2$

$r^2=1+\frac{9}{4}$

$r^2=\frac{4+9}{4}$

$r^2=\frac{13}{4}$

$r=\sqrt{\frac{13}{4}}$

$r=\frac{\sqrt{13}}{2}$

Therefore, Equation of Circle is $(x-1)^2+(y-\frac{3}{2})^2=\frac{13}{4}$