Sorry make a typo mistake.
Question Should be
Find Coordinates of G instead.
Hi Guys !
I am currently stuck with this question :
A Circle and the curve y^2=kx have a common tangent at the point B. The Centre G of the circle lies on the Line AC. Find
i) The Coordinates of C
ii) equation of the circle
Whereby A = (1,-2) ; B = (4,4) and C = (9,6) and k = 4.
Need some guidance in this question !
Realli killing tons of my brain cells thinking about this question man !
Appreciate your help
Thanks !
Firstly, formulate the fact that the centre of the circle (h, k) is on AC.
Therefore, write the equation of AC by first calculating the slope,
then using A or C to write the equation.
Replace x and y with h and k.
This relates the centre co-ordinates.
You need 2 equations to solve for 2 unknowns, hence a second relationship is needed.
The centre is also on the line going from B to (h, k).
It's slope is perpendicular to the tangent.
Hence, first write the slope of the tangent and write the equation of the 2nd centreline in terms of h and k.
So you want and to have a common tangent at (4, 4). One thing that means is that each curve must go through (4, 4). In particular, so k= 4 as you were apparently given, and the parabola is given by . Then and at y= 4, that is or .
In order that the circle , which has derivative given by , have the same tangent at (4, 4) we must have as well as or . The line through AC, that is, the line through (1, -2) and (9, 6) has slope (6-(-2))/(9- 1)= 8/8= 1 so the line itself is y= 1(x-1)-2= x- 3. Since the center of the circle lies on that line, we must have .
That is, we have three equations, , , and . It should be easy to solve the last two linear equations for and . You could, if you wish, put those values into the first equation and solve for R but apparently the problem only asks for the coordinates of the center of the circle.
Problem is simplified by using the formula for the equation of tangent to a parabola at a given point
parabola Y^2 =4ax
equation of tangent=2ax/y1 +2ax1/y1 at x1,y1
bjh
Maybe simplest is to use the slope of the tangent to get the equation of a 2nd centreline.
(for the first equation see earlier post)
At the point B(4,4), the tangent slope is
which means that the perpendicular circle centreline passing through (4,4) has slope and contains (h,k).
Use to write the equation, so .
Solving the simultaneous equations (centreline equations) finds the centre co-ordinates.