To answer that question we will use the general circumference equation, and the established condition in each case
That is: ( x - h )² + ( y - k )² = r²
Solutions are:
a) x² + ( y - 5 )² = 25
b) (x - 1)² + ( y - 4 )² = 2
c)( x + 5 )² + ( y - 4 )² = 16
d)( x - 3 )² + ( y - 5 )² = 9
In the equation : ( x - h )² + ( y - k )² = r²
r is radius of the circumference C ( h ; k ) are center coordinates
a) C ( 0 ; 5 ) contains P ( 0;0)
Then |CP| is radius CP = [ 0-0 ; 0-5 ] CP = ( 0 ; -5 )
|CP| = √ 0 + (-5)²
|CP| = r = 5
Then ( x - h )² + (y - k )² = r² becomes
( x - 0 )² + ( y - 5 )² = (5)²
x² + ( y - 5 )² = 25
b) Diameter endpoints P ( 2 ; 5 ) Q ( 0 ; 3)
PQ = [ 0 - 2 ; 3 - 5 ) = [ -2 ; -2 ]
|PQ| = √(-2)² + (-2)² = 2×√2
then radius is |PQ|/2 r = √2
Midle point ( center ) C = [ (x₁ + x₂)/2 : y₁ + y₂)/2]
C = [ (2 + 0) /2 ; ( 5+ 3 )/2 ]
C = ( 1 ; 4 )
Then
(x - 1)² + ( y - 4 )² = 2
c) Tangent to x-axis r = 4 and point ( -5 ; 8 )
From attached file ( C) we can see:
C ( -5 ; 4 ) and diameter endpoints are P ( -5 0 ) and Q ( -5 ; 8 )
The equation is :
( x -(-5) )² + ( y - 4 )² = 16
( x + 5 )² + ( y - 4 )² = 16
d)
Center on x = 3 tangent to y-axis at P ( 0;5)
If the circumference is tangent to y-axis then r = 3
Again from the attached file (D) the red line mark two endpoints diameter
C = ( 3 ; 5 )
Then the equation is:
( x - 3 )² + ( y - 5 )² = 9
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