how do i write the equation of the circle in standard form. I have to find the center,radius and intercept then graph the circle. x^2+y^2-4x+18y=-69
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To find the coordinates of the center complete the square for both x and y.
To do this halve the coefficients in front of the terms in x and y which gives -2 and 9 respectively. Considering the expressions
`(x-2)^2 = x^2 - 4x + 4` and `(y+9)^2 = y^2 + 18y + 81`
we can see that the first 2 terms on the lefthand sides of these two equations give us the terms in `x^2, x, y^2` and `y` in the original equation
`x^2 + y^2 - 4x +18y = -69`
Rearranging the expressions,
`(x-2)^2 - 4 = x^2 - 4x` and `(y+9)^2 - 81 = y^2 + 18y`
Substituting these two expressions into the original equation gives
`(x-2)^2 - 4 + (y+9)^2 - 81 = -69`
Collect all the integer terms onto the righthand side, giving
`(x-2)^2 + (y+9)^2 = -69 + 4 + 81`
So we have
`(x-2)^2 + (y+9)^2 = 16` in standard form answer
At any point `(x,y)` on the circle we can form a right-angled triangle from that point to the centre of the circle `(x_c, y_c )` .
The hypotenuse is the radius of the circle and the horizontal and vertical sides have length `(x-x_c)` and `(y-y_c)` respectively
So using Pythagoras' Theorem `(x-x_c)^2 + (y-y_c)^2 = r^2`
This is exactly the standard form we have obtained above. We can see that setting
`x_c = 2` , `y_c = -9` and `r^2 = 16`
So the centre is at `(x_c,y_c) = (2, -9)` and the radius is `r = sqrt(16) = 4`
answer
To draw the graph on a computer, convert to polar coordinates
`x(theta) = rcos(theta) = 4cos(theta)`
`y(theta) = rsin(theta) = 4sin(theta)`
` `where `theta` is the angle the hypotenuse (radius) makes from the horizontal line
`y = y_c = -9`
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