# Thread: New! coordinate Geometry

1. ## New! coordinate Geometry

Could someone show me how to work out these please

1. The circle x^2(y+3)^2=25 meets the x axis at p and q, find the coordinates of p and q?

2. The points A(7,4) B(8,-7) and C(-4,3) lie on a circle. Find the equation of the circle?

3. Find the equation of the tangent at the point (3,1) on the circle
x^2+y^2-4x+10y-8=0

Thanks for your help

2. Originally Posted by Colin_m
1. The circle x^2(y+3)^2=25 meets the x axis at p and q, find the coordinates of p and q?
$x^2(y+3)^2=25$ isn't a circle. Write it properly.

Originally Posted by Colin_m
2. The points A(7,4) B(8,-7) and C(-4,3) lie on a circle. Find the equation of the circle?
Plug these points in $(x-a)^2 + (y-b)^2 = r^2$ and find a, b and r.

Originally Posted by Colin_m
3. Find the equation of the tangent at the point (3,1) on the circle
x^2+y^2-4x+10y-8=0
Use implicit differentiation. Or write the circle as two functions of x, then use one of them and find the slope. Then use the point given to find the equation.

3. thanks.
isn't a circle. Write it properly.
my maths teacher is think then lol

4. Hello, Colin!

2. The points $A(7,4),\: B(8, \text{-}7),\: C(\text{-}4,3)$ lie on a circle.
Find the equation of the circle.
On a hunch, I found the slopes of the sides of this triangle.

$m_{AB} \:=\:\frac{\text{-}7-4}{8-7} \:=\:-11$

$m_{AC} \:=\:\frac{3-4}{\text{-}4-7} \:=\:\frac{1}{11}$

Hence: . $AB \perp AC\quad\hdots \text{ We have a right triangle!}$

The center is the midpoint of the hypotenuse $BC.$

. . Center: . $O\left(\frac{8-4}{2},\:\frac{-7+3}{2}\right) \;=\;(2,-2)$

The radius is the length of $OA$ (or $OB$ or $OC$).

. . $r^2 \;=\;OA^2 \;=\;(7-2)^2 + (4+2)^2 \;=\;61$

The equation is: . $(x-2)^2 + (y+2)^2 \:=\:61$