Question

what is the radius of the circle shown below? x^2+y^2-8x-10y-8=0

Answer 1

Do u know circle equation

Answer 2

To answer questions 1 and 2, you need to find a, b and r such that:

(x-a)² + (y-b)² = r²
⇔ x² - 2ax + y² - 2by + a² + b² - r² = 0

So for this equation:

x^2 + y^2 -8x -10y -8=0

we can match up coefficients:

-2a = -8
-2b = -10
a² + b² - r² = -8

Solving this system of equations gives:

a = 4
b = 5
r = +/-7

By convention, we take r as positive. So the centre of the circle is at (4,5) and the radius is 7.

For question 3, remember that the equation of an aligned ellipse is:

(x-c)²/a² + (y-d)²/b² = 1

where (c,d) is the centre and a and b are the lengths of the half-axes. In your equation, a=7 and b=5, so the major half-axis has length 7, and so the major axis has length 14.

Question 4 you can do yourself. Just plug in the values of x and y and see which ones work. For example, for A, plugging x = -11 and y = 1 into the equation of the conic section gives:

(-11 + 2)²/64 + (1-1)²/81 = 81/64 ≠ 1

This doesn't satisfy the equation, so A is incorrect.

In question 5, I suggest shifting the figure to the origin, which doesn't change the length of the transverse axis:

y^2/16 - x^2/9 = 1

The equation is that of a hyperbola. The transverse axis is the one with the positive sign, which in this case is the y axis. So to find the points where the conic section intersects the y axis, set x=0 and solve:

y^2/16 = 1
⇔ y² = 16
⇔ y = +/- 4

The two solutions are the points (0,-4) and (0,4). The distance between these two points is 8, which is the length of the transverse axis. got it from yahoo answers

Answer 3

no problem :)

Answer 4

awesome :D

Answer 5

yahoo answers are the best and that's where I got it from:)