circle equation part2

Quote:

Originally Posted by bigmayne93

A circle has equation x^2+y^2-4x+4y-12=0

a)i)find the co-ordinates of the centre of the circle
ii)the radius of the circle

b)find the co-ordinates of the two points where the circle crosses the x axis

c)find the equation of the tangent to the circle at point (4,2)

Thanks to anyone who can help (Happy)

Draw a sketch!

to a) (i), (ii)
The equation of a circle with the center $M\left(x_M, y_M\right)$ is:

$(x-x_M)^2+(y-y_M)^2=r^2$

Use completing the square to transform the given equation into the general form:

$x^2-4x\bold{\color{red}+4}+y^2+4y\bold{\color{blue}+4} =12 \bold{\color{red}+4} \bold{\color{blue}+4}$

to b)

If the circle crosses the x-axis then y = 0. Plug in this value into the given equation and solve for x.

to c)

The tangent is perpendicular to the line MP. Calculate the slope of MP, calculate the perpendicular slope, which is the slope of the tangent, use point-slope-formula of a straight line to get the equation of the tangent.