What are the solutions to the equation (22-r)^2+(11-r)^2=r^2 ?

The solutions to the equation (22-r)^2+(11-r)^2=r^2 are r=55,r=11

Find the zeros of (22-r)^2+(11-r)^2=r^2

The zeros of (22-r)^2+(11-r)^2=r^2 are r=55,r=11

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(22 - r)^{2} + (11 - r)^{2} = r^{2}

r = 55, r = 11

Solution steps

(22 - r)^{2} + (11 - r)^{2} = r^{2}

Expand

(22 - r)^{2} + (11 - r)^{2} : 2 r^{2} - 66 r + 605

2 r^{2} - 66 r + 605 = r^{2}

Move

r^{2}

to the left side

r^{2} - 66 r + 605 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 66 ) \pm ( - 66 ) ^{2} - 4 \cdot 1 \cdot 605}{2 \cdot 1}

(- 66)^{2} - 4 \cdot 1 \cdot 605 = 44

r_{1, 2} = \frac{- ( - 66 ) \pm 44}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 66 ) + 44}{2 \cdot 1}, r_{2} = \frac{- ( - 66 ) - 44}{2 \cdot 1}

r = \frac{- ( - 66 ) + 44}{2 \cdot 1} : 55

r = \frac{- ( - 66 ) - 44}{2 \cdot 1} : 11

The solutions to the quadratic equation are:

r = 55, r = 11

5 x^{2} + 7 x = 3

(p - 3)^{2} - 4 (p - 3)^{2} + 3 = 0

4 x + x^{2} = 21

x^{2} + 3 x - 1 = 0

12 x^{2} + 25 x + 7 = 0

What are the solutions to the equation (22-r)^2+(11-r)^2=r^2 ?
The solutions to the equation (22-r)^2+(11-r)^2=r^2 are r=55,r=11
Find the zeros of (22-r)^2+(11-r)^2=r^2
The zeros of (22-r)^2+(11-r)^2=r^2 are r=55,r=11