circle equation part2

• Nov 3rd 2009, 10:48 AM
bigmayne93
circle equation part2
A circle has equation x^2+y^2-4x+4y-12=0

a)i)find the co-ordinates of the centre 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)
• Nov 3rd 2009, 12:06 PM
earboth
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

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 $\displaystyle M\left(x_M, y_M\right)$ is:

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

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

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

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.