# Thread: Coordinate Geometry of Circles

1. ## Coordinate Geometry of Circles

The equation of a circle, C is x^2+y^2-8x+10y+28=0

(i) Find the coords. of the centre of C and find the radius of C.

(ii)Show that the line L:2x+3y=6 is a tangent to the circle and find the coords. of the point of contact.

Please do part(ii). Anyway i have found the coords. of the pt. of contact which is (6,-2). But how to show that the line is a tangent to the circle??

2. Originally Posted by maybeline9216
The equation of a circle, C is x^2+y^2-8x+10y+28=0

(i) Find the coords. of the centre of C and find the radius of C.

(ii)Show that the line L:2x+3y=6 is a tangent to the circle and find the coords. of the point of contact.

Please do part(ii). Anyway i have found the coords. of the pt. of contact which is (6,-2). But how to show that the line is a tangent to the circle??
The line from point of contact to centre is radius. If this line (radius) is perpendicular to the given line 2x + 3y =6, then this given line is tangent.

Now
Centre = (4, -5)
Point of contact=(6, -2)

slope of radius $\displaystyle = \frac{-2-(-5)}{6-4}=\frac{3}{2}$

Slope of given line 2x + 3y = 6 is $\displaystyle = \frac{-2}{3}$

Product of their slopes $\displaystyle = \frac{3}{2} \times \frac{-2}{3} = -1$

So, the given line is perpendicular to the radius at the point of contact.
so, it is a tangent at rhe point of contact.