# Math Help - circle analytic geometry

1. ## circle analytic geometry

Hello there. I'm currently studying AS level maths in the UK and have been given a sheet of circle questions (This is not my strong point either).
I'm hoping you can help with this .

1) A circle passes through the points (2,0) and (8,0) and has the y axis as a tangent. Find the two possible equations.

2) A(6,3) and B(10,1) are two points on a circle with centre (11,8).
Calculate
Part 1 - the co-ordinates of the mid-point of the chord
Part 2 - the disbance of the chord AB from the centre of the circle
Part 3 - the radius of the circle.

2. For question one:
Let the center of the circle be (h, k).
Since the circle is tangent to the y-axis, the point (0, k) is on the circle.
And since both (2,0) and (8,0) are on the circle, they form an isosceles triangle with (h, k), which means that h = 5. Hence, (h, k) = (5, k). So the radius of the circle is 5.

Solving for k by using the distance formula:
$\sqrt{(5-2)^2 + (k-0)^2} = 5$

$\sqrt{3^2 + k^2} = 5$

$9 + k^2 = 25$

$k^2 = 16$

$k = \pm4$.

Hence the two equations are:

$(x - 5)^2 + (y - 4)^2 = 25$

$(x - 5)^2 + (y + 4)^2 = 25$

3. Thanks so much for your help, actually understand that question well. Now the other question :S

4. the chord is just the line connecting the two given points

for two points $(x_1,y_1)$ and $(x_2,y_2)$

the midpoint of them would be $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

so the point for part one is $(\frac{6+10}{2},\frac{3+1}{2})=(8,2)$

parts two and three are just applying the distance formula, the first one is finding the distance from the center point (given) to the point I found in part 1 and the second is using it to find the distance between the center and any one of the two points they've given you

the distance formula is
$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

can you do it now?