What are the solutions to the equation solvefor x,x^2+y^2=2cx ?

The solutions to the equation solvefor x,x^2+y^2=2cx are x=c+sqrt(-y^2+c^2),x=c-sqrt(-y^2+c^2)

Find the zeros of solvefor x,x^2+y^2=2cx

The zeros of solvefor x,x^2+y^2=2cx are x=c+sqrt(-y^2+c^2),x=c-sqrt(-y^2+c^2)

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solve for

x, x^{2} + y^{2} = 2 c x

x = c + - y^{2} + c^{2}, x = c - - y^{2} + c^{2}

Solution steps

x^{2} + y^{2} = 2 c x

Move

2 c x

to the left side

x^{2} + y^{2} - 2 c x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 2 c x + y^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 c ) \pm ( - 2 c ) ^{2} - 4 \cdot 1 \cdot y ^{2}}{2 \cdot 1}

Simplify

(- 2 c)^{2} - 4 \cdot 1 \cdot y^{2} : 2 c^{2} - y^{2}

x_{1, 2} = \frac{- ( - 2 c ) \pm 2 c ^{2} - y ^{2}}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 c ) + 2 c ^{2} - y ^{2}}{2 \cdot 1}, x_{2} = \frac{- ( - 2 c ) - 2 c ^{2} - y ^{2}}{2 \cdot 1}

x = \frac{- ( - 2 c ) + 2 c ^{2} - y ^{2}}{2 \cdot 1} : c + - y^{2} + c^{2}

x = \frac{- ( - 2 c ) - 2 c ^{2} - y ^{2}}{2 \cdot 1} : c - - y^{2} + c^{2}

The solutions to the quadratic equation are:

x = c + - y^{2} + c^{2}, x = c - - y^{2} + c^{2}

x^{2} + 6 x + 1 = - 2 x + 1 - 2 x + 1

x^{2} - 81 = (x - 9) 2

x, x^{2} - 4 y^{2} - 8 y = 4

x, 2 x^{2} + 5 y^{2} + z^{2} = 4

u^{2} - 8 u = 0

What are the solutions to the equation solvefor x,x^2+y^2=2cx ?
The solutions to the equation solvefor x,x^2+y^2=2cx are x=c+sqrt(-y^2+c^2),x=c-sqrt(-y^2+c^2)
Find the zeros of solvefor x,x^2+y^2=2cx
The zeros of solvefor x,x^2+y^2=2cx are x=c+sqrt(-y^2+c^2),x=c-sqrt(-y^2+c^2)