# Find the radius of a circle with the equation: x^2+y^2-14y+124=0?

geometry

### 9 Answers

- KrishnamurthyLv 76 months ago
x^2 + y^2 - 14y + 124 = 0

This isn't a circle.

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- Φ² = Φ+1Lv 76 months ago
√(0²+(-7)²-124) = √-75 which is an invalid value for a radius on the real plane.

Source(s): x² + y² + 2ax + 2by + c = 0 rearranges to (x+a)² + (y+b)² = a² + b² - c, a circle centred at (-a,-b) with a radius of √(a²+b²-c)- Log in to reply to the answers

- Ian HLv 76 months ago
x^2 + y^2 - 14y + 124 = 0

The equation you gave, (above), was not a circle. Perhaps it should have been

x^2 + y^2 - 14y = 124

x^2 + y^2 - 14y + 49 = 124 + 49

x^2 + (y - 7)^2 = 173

Centre (0, 7) radius √(173)

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- PinkgreenLv 76 months ago
Ans. such circle does not exist, check out

your problem.

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- lenpol7Lv 76 months ago
Complete the square in 'x' & 'y'

Hence

x = 0

(y - 14/2)^2 - (14/2)^2 = 0

( y - 7)^2 = 196/4

y - 7 = 14/2 = 7

y = 14

Hence 124 = 0^2 + 14^2=

124 = 49

=75 which is r^2

r = sqrt(75) = 5sqrt(3) or 8.660....

Hence the original eq'n is

(x - 0)^ + ( y - 14)^2 = 75

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- Mike GLv 76 months ago
x^2+y^2-14y+49 = -124+49

x^2+(y-7)^2 = -75

Radius is an imaginary number

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- DWReadLv 76 months ago
The equation you've provided is not a circle. Did you mean to write

x² + y² - 14y - 124 = 0?

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- llafferLv 76 months ago
If that's supposed to be:

x² + y² - 14y - 124 = 0

then we can move the constant to the right side:

x² + y² - 14y = 124

Then complete the square with the y's adding 49 to both sides:

x² + y² - 14y + 49 = 124 + 49

Now factor the y's as a perfect square trinomial and simplify the right side:

x² + (y - 7)² = 173

The center of this is (0, 7) and the radius is √173

Your equation as shown is not a circle.

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- 6 months ago
x^2 + y^2 - 14y = -124

x^2 + y^2 - 2 * 7 * y = -124

x^2 + y^2 - 2 * 7 * y + 49 = 49 - 124

(x - 0)^2 + (y - 7)^2 = -75

This isn't a circle.

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