First, convert the polar equation into rectangular. To do this multiply both sides by r to get:
r^2 = ar*sinθ + br*cosθ.
Using r^2 = x^2 + y^2, y = r*sinθ, and x = r*cosθ, this becomes:
x^2 + y^2 = ay + bx
==> (x^2 - bx) + (y^2 - ay) = 0, by bringing all the x and y-terms over
==> (x^2 - bx + b^2/4) + (y^2 - ay + a^2/4) = (a^2 + b^2)/4, by completing the square
==> (x - b/2)^2 + (y - a/2)^2 = (a^2 + b^2)/4 = [√(a^2 + b^2)/2]^2.
This shows that the conic is a circle with center (b/2, a/2) and radius √(a^2 + b^2)/2.
I hope this helps!
x2 + y2 + 8x - 4y = -11 x^2 + 8x + y^2 -4y = -11 x^2+8x +16-16 +y^2-4y+4-4 = -11 (x+4)^2 -16 +(y-2)^2 -4 = -11 (x+4)^2 +(y-2)^2 = -11+4+16 (x+4)^2 +(y-2)^2 = 9 Equation of a circle with middle (-4,2) and radius 3.
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First, convert the polar equation into rectangular. To do this multiply both sides by r to get:
r^2 = ar*sinθ + br*cosθ.
Using r^2 = x^2 + y^2, y = r*sinθ, and x = r*cosθ, this becomes:
x^2 + y^2 = ay + bx
==> (x^2 - bx) + (y^2 - ay) = 0, by bringing all the x and y-terms over
==> (x^2 - bx + b^2/4) + (y^2 - ay + a^2/4) = (a^2 + b^2)/4, by completing the square
==> (x - b/2)^2 + (y - a/2)^2 = (a^2 + b^2)/4 = [√(a^2 + b^2)/2]^2.
This shows that the conic is a circle with center (b/2, a/2) and radius √(a^2 + b^2)/2.
I hope this helps!
x2 + y2 + 8x - 4y = -11 x^2 + 8x + y^2 -4y = -11 x^2+8x +16-16 +y^2-4y+4-4 = -11 (x+4)^2 -16 +(y-2)^2 -4 = -11 (x+4)^2 +(y-2)^2 = -11+4+16 (x+4)^2 +(y-2)^2 = 9 Equation of a circle with middle (-4,2) and radius 3.