# Help Need in Help With finding centre of circle

• Nov 28th 2010, 06:59 AM
liukawa
Help Need in Help With finding centre of circle
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 !

Thanks !
• Nov 28th 2010, 07:00 AM
liukawa
Sorry make a typo mistake.

Question Should be

• Nov 28th 2010, 08:05 AM
Quote:

Originally Posted by liukawa
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 !

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.
• Nov 28th 2010, 08:10 AM
HallsofIvy
So you want $y^2= kx$ and $(x- x_0)^2+ (y- y_0)^2= R^2$ to have a common tangent at (4, 4). One thing that means is that each curve must go through (4, 4). In particular, $4^2= 16= k(4)$ so k= 4 as you were apparently given, and the parabola is given by $y^2= 4x$. Then $2y y'= 4$ and at y= 4, that is $8y'= 4$ or $y'= \frac{1}{2}$.
In order that the circle $(x- x_0)^2+ (y- y_0)^2= R^2$, which has derivative given by $2(x- x_0)+ 2(y- y_0)y'= 0$, have the same tangent at (4, 4) we must have $(4- x_0)^2+ (4- y_0)^2= R^2$ as well as $2(4- x_0)+ 2(4- y_0)\frac{1}{2}= 0$ or $2x_0+ y_0= 12$. 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 $y_0= x_0- 3$.

That is, we have three equations, $(4- x_0)^2+ (4- y_0)^2= R^2$, $2x_0+ y_0= 12$, and $y_0= x_0- 3$. It should be easy to solve the last two linear equations for $x_0$ and $y_0$. 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.
• Nov 28th 2010, 12:34 PM
bjhopper
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
• Nov 28th 2010, 04:21 PM
$y^2=4x\Rightarrow\ 2y\frac{dy}{dx}=4\Rightarrow\ \frac{dy}{dx}=\frac{4}{2y}$
At the point B(4,4), the tangent slope is $\frac{4}{(2)4}=\frac{1}{2}$
which means that the perpendicular circle centreline passing through (4,4) has slope $-2$ and contains (h,k).
Use $y-y_1=m\left(x-x_1\right)$ to write the equation, so $k-4=-2(h-4)$.