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Popular >

p^2+q^2=1

Solution

p^{2} + q^{2} = 1

Solution

(a, b) = (0, 0), r = 1

Solution steps

p^{2} + q^{2} = 1

Rewrite

p^{2} + q^{2} = 1

in the form of the standard circle equation

(p - 0)^{2} + (q - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 1

Popular Examples

x^2+y^2+14x-6y-6=0

x^{2} + y^{2} + 14 x - 6 y - 6 = 0

x^2+y^2-2x+14y+34=0

x^{2} + y^{2} - 2 x + 14 y + 34 = 0

x^2-2x+1+y^2+y-4/4 = 11/4

x^{2} - 2 x + 1 + y^{2} + y - \frac{4}{4} = \frac{11}{4}

(x-5)^2+(y+7)^2=4

(x - 5)^{2} + (y + 7)^{2} = 4

(x-1)^2+(y-(-1/2))^2=2^2

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

Frequently Asked Questions (FAQ)

What is p^2+q^2=1 ?
The solution to p^2+q^2=1 is Circle with (a,b)=(0,0),r=1