Can pls. help me this q:how to find the equation of a circle

Question

passing through (4,-3) and (-3,-4) with radius 5? i need it now?  

 

 

Answer 1

Points are (4, -3) and (-3, -4) and the radius of the circle as 5.

General Equation of the circle is (x-h)²+(y-k)² = r² where r is the radius of the circle.

Point (4, -3) and r = 5.

(4 - h)² + (-3 - k)² = 5²

(4 - h)² + (3 + k)² = 25

16 + h² - 8h +9 + k² + 6k = 25

h² + k² - 8h + 6k = 0.             (Name it as Equation 1)

Point (-3, -4) and r = 5.

(-3 - h)² + (-4 - k)² = 5²

(3 + h)² + (4 + k)² = 25

9 + h² + 6h +16 + k² + 8k = 25

h² + k² + 6h + 8k = 0.             (Name it as Equation 2)

Solving the Equations.

Take Equation 1.

h² + k²  = 8h - 6k

Substitute it in Equation 2 and simplify.

8h - 6k + 6h + 8k = 0

14h + 2k = 0

k = -7h.

Substitute h = -k in Equation 1.

h² + (-7h)² - 8h + 6(-7h) = 0

50h² - 50h= 0

h = 0, 1

Then k = 0, -7                        (since h = -k)

The points are (0,0) and (1,-7).

The Equation of the circle with centre (0,0) is a x² + y² = 25

The Equation of the circle with centre (1,-7) is a point circle is (x - 1)² + (y + 7)² = 25

Therefore Equation of the circle is (x - 1)² + (y + 7)² = 25.