# The equation of circle passing through (4,5) and having centre (2,2)

Arun
25757 Points
5 years ago
1. (4, 5) lies on the circle. Use the distance formula to find the distance from the center, (2, 2) to (4, 5). That would be the radius.

d = √[ (x₂ - x₁)² + (y₂ - y₁)² ]
r  = √[ (4 - 2)² + (5 - 2)² ]
r  = √13

The equation of a circle in standard form is
(x - a)² + (y - b)² = r²          center = (a, b)

(x - 2)² + (y - 2)² = (√13)²
(x - 2)² + (y - 2)² = 13

2. You need to find the equation of a circle with the same center as x² + y² - 4x - 6y + 9 = 0.

x² + y² - 4x - 6y + 9 = 0
(x² - 4x) + (y² - 6y) = -9

To complete the square, take half of 4 and half of 6, then square it and add it to both sides.
(4/2)² = 4
(6/2)² = 9

(x² - 4x + 4) + (y² - 6y + 9) = -9 + 4 + 9
(x - 2)(x - 2) + (y - 3)(y - 3) = 4
(x - 2)² + (y - 3)² = 2²

center = (2, 3)

To be concentric, the circle we're looking for also has a center at (2, 3). Since (-4, 5) lies on this circle, we can find the radius by finding the distance between those points.

r = √[ (-4 - 2)² + (5 - 3)² ]
r = √40

(x - a)² + (y - b)² = r²          center = (a, b)
(x - 2)² + (y - 3)² = (√40)²
(x - 2)² + (y - 3)² = 40

3.
(1, -2) lies on the circle.
(1 - a)² + (-2 - b)² = r²

(4, -3) also lies on the circle.
(4 - a)² + (-3 - b)² = r²

Since both equations have the same radius, they are equal to each other.
(1 - a)² + (-2 - b)² = (4 - a)² + (-3 - b)²
1 - 2a + a² + 4 + 4b + b² = 16 - 8a + a² + 9 + 6b + b²
a² - a² + b² - b² - 2a + 8a + 4b - 6b + 1 + 4 - 9 - 16 = 0
6a - 2b - 20 = 0

(a, b) is the center. Since the center lies on the line 3x + 4y = 7, the following holds true:
3a + 4b = 7

6a - 2b - 20 = 0
6a - 20 = 2b
(6a - 20) / 2 = b
3a - 10 = b

3a + 4b = 7
3a + 4(3a - 10) = 7
3a + 12a - 40 = 7
15a = 47
a = 47/15
a = 3.133

3a - 10 = b
3(47/15) - 10 = b
-3/5 = b
-0.6 = b

center = (a, b) = (3.133, -0.6)

Pick one of the points to find the radius. I'll use (1, -2).
r = √[ (1 - 3.133)² + (-2 - (-0.6))² ]
r = √6.51