1. ## The Circle?

"A circle passes through the origin (0,0) and the points (4,0) and (3,3).
Find:
A. The centre of the circle
B. The equation to the tangent of the circle to the origin"

Cheers

2. Originally Posted by dungo
"A circle passes through the origin (0,0) and the points (4,0) and (3,3). Find:
A. The centre of the circle
B. The equation to the tangent of the circle to the origin"
Find the perpendicular bisectors of the two line segments with endpoints (0,0)&(4,0) and (0,0)&(3,3).
Find where those two lines intersect, that is the center of the circle.

3. Originally Posted by dungo

"A circle passes through the origin (0,0) and the points (4,0) and (3,3).
Find:
A. The centre of the circle
B. The equation to the tangent of the circle to the origin"

Cheers
The general equation of a circle is

$x^2+y^2+2fx+2gy+c=0$, where f, g and c are constants.

Put first point (0,0) in eqn, we got,

0 + 0 + 0 + 0 + c = 0

$\Rightarrow c = 0$

Now, the eqn becomes,

$x^2+y^2+2fx+2gy=0$

Put second point (4,0) in eqn, we got,

$(4)^2+(0)^2+2f(4)+2g(0)=0$

$\Rightarrow f = -2$

Now, the eqn becomes,

$x^2+y^2-4x+2gy=0$

Put third point (3,3) in eqn, we got,

$(3)^2+(3)^2-4(3)+2g(3)=0$

$\Rightarrow g = -1$

Now, the eqn becomes,

$x^2+y^2-4x-2y=0$

Complete the square for x and y terms,

$[x^2-2(2)x+2^2]-4+[y^2-2(1)y+1^2]-1=0$

$(x-2)^2+(y-1)^2-4-1=0$

$(x-2)^2+(y-1)^2=5$

so, centre = (2, 1) and radius $= \sqrt {5}$

B) Now, to find the slope of tangent, we can find the slope of the normal through centre (2, 1) to (0, 0)

slope of normal $= \frac{1-0}{2-0}=\frac{1}{2}$

so, slope of tangent = -2

now, find the equation of tangent line through (0, 0) having slope -2,

y = mx + b

0 = -2(0) + b

b = 0

so, eqn of tangent is

y = -2x

4. Hello, dungo!

Did you make a sketch?

A circle passes through the origin (0,0) and the points A(4,0) and B(3,3).

Find:
A. The centre of the circle
B. The equation to the tangent of the circle to the origin
Code:
        |       : B
|       : o (3,3)
|       *  *
|     * :   *
|   *   :    *
| *     :     *
- * - - - + - - -o - - -
|O      :    (4,0)
|       :      A

Since $O$ and $B$ are on the x-axis, we already know the perpendicular bisector of $OA.$
. . It is the vertical line: $x = 2$

Now we seek a point on that vertical line $P(2,y)$
. . which is equidistant from $O$ and $B$

$OP^2 \:=\:2^2 + y^2\:=\:y^2 + 4$
$BP^2 \:=\:(2-3)^2 + (y-3)^2 \:=\: y^2 - 6y + 10$
. . We have: . $y^2+4 \:=\:y^2 - 6y + 10 \quad\Rightarrow\quad y \:=\:1$

Therefore, the center is $(2,1)$

You should be able to complete the problem now . . .