circles equations, complete the square?

Question

complete the square of circle with centre c x^2 + y^2 -8x=6y=11 in the form (x-a)^2 + (y-b)^2 = r^2

then find the coordinates of c and the radius of the circle

c0 then o point has coordinates (0,0)
find teh length of co and determine whether point o lies inside or outside the circle with a reason

+ 6y sorry for my mistake

Answer 1

Not sure if you meant -6y or +6y

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x^2 + y^2 -8x-6y=11
x^2 -8x +16 + y^2 -6y +9 = 11 + 16 + 9
(x-4)^2 + (y-3)^2 = 36

Center is (4,3)
Radius is 6

Distance from (0,0) to (4,3) is sqrt(4^2 + 3^2) or 5
5 is less than the radius 6, so the point is IN the circle.
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x^2 + y^2 -8x+6y=11
x^2 -8x +16 + y^2 +6y +9 = 11 + 16 + 9
(x-4)^2 + (y+3)^2 = 36

Center is (4,-3)
Radius is 6

Distance from (0,0) to (4,-3) is sqrt(4^2 + (-3)^2) or 5
5 is less than the radius 6, so the point is IN the circle.

Answer 2

it is an ellipse, no longer a circle. How can i tell? via fact, as you rightly observed, the coefficients on the squared words are no longer an identical. yet i'm going to allow you to anyhow. 2x^2 + 3x + y^2 + 4y = 0 finished the sq. for the y's first by way of including 4 to the two factors. 2x^2 + 3x + y^2+4y+4= 4 2x^2 + 3x + (y+2)^2 = 4 multiply the two factors by way of 2 to make the coefficient on the x^2 a suited sq. (4) 4x^2 + 6x + 2(y+2)^2 = 8 enable's see... how would we finished the sq. for the x's? (2x+ok)(2x+ok) = 4x^2 + 4kx + ok^2 = 4x^2 + 6x + ok^2 so 4k=6 so ok=6/4 = 3/2 so ok^2 is 9/4; it is what we would desire to continuously upload to the two factors of that equation from previously. 4x^2+6x+ 9/4 + 2(y+2)^2 = 8 + 9/4 = 32/4 + 9/4 = 40-one/4 enable's multiply each little thing by way of 4, via fact no one likes fractions! 16x^2+24x+9 + 8(y+2)^2 = 40-one factor the x area... (4x+3)^2 + 8(y+2)^2 = 40-one we are in a position to opt for whilst we'd opt for to be accomplished with this.... it is purely algebra in spite of each little thing :) divide the two factors by way of sixteen so as that the coefficient on x is a million. (x + 3/4)^2 + (a million/2)(y+2)^2 = 40-one/sixteen. so i can actual show you how to comprehend that the middle of the ellipse is x = -3/4 and y = -2 additionally, the horizontal (x) "radius" would be (sqrt(40-one))/4, on an identical time as the vertical (y) "radius" would be double that (via fact there's a a million/2 in front of the y^2 area). So this ellipse sounds like an egg... it is taller than that's huge.

Answer 3

x² + y² - 8x + 6y = 11
(x² - 8x + 16) + (y² + 6y + 9) = 36
(x - 4)² + (y + 3)² = 6²
Centre (4 , - 3)
Radius = 6
CO = 5
Point O lies inside circle.