1. ## Co-ordiante geometry

Can anyone help me with these with step by step answers.

Here are a few of them . . .

You're expected to know the general equation of the circle.
. . The circle with center C(h,k) and radius r has the equation:
. . . . (x - h)² + (y -k)² .= .

1) Find the coordinates of the center and the radius of the circle:
. . . . . (x - 7)² + (y + 2)² .= .64
Do you know how to use that general equation?

The center is (7,-2) and the radius is 8.

2) The points P(10,-4) and Q(2,2) are the ends of a diameter.
. . Find the equation of the circle.
The center is the midpoint of diameter PQ.

We have: .x = ½(10 + 2) = 6 .and .y = ½(-4 + 2) = -1
. . Hence, the center is: C(6,-1)

The radius is the distance CP or CQ.
. . CQ .= .√[(2 - 6)² + (2 + 1)²] .= .√[16 + 9] .= .√[25] .= .5

Therefore, the equation is: .(x - 6)² + (y + 1)² .= .25

3) The point (-2,3) lies on the circle: .(x - 4)² + (y + 1)² .= .
. . Find the exact value of r.

If (-2,3) lies on the circle, then x = -2, y = 3 satisfies the equation.

We have: .(-2 -4)² + (3 + 1)² .= . . . .= .(-6)² + 4² .= .36 + 16 .= .52

. . . . . . - . - . . . .__ . . . . .__
Therefore: .r .= .√52 .= .2√13

3. Thanks but has anyone got the answer to the rest of the questions.

Can anyone help me with these with step by step answers.
4. (a) p(-1,2), C(4,5)

Equation of a circle is of the form:

(x - h)^2 + (y - k)^2 = r^2
where (h,k) is the center of the circle and r is the radius
using (x,y) = (-1,2) and (h,k) = (4,5)
=> (-1 - 4)^2 + (2 - 5)^2 = r^2
=> 25 + 9 = r^2
=> 34 = r^2
=> r = sqrt(34)

thus the equation of the circle is:
(x - 4)^2 + (y - 5)^2 = 34

(b) (x - 4)^2 + (y - 5)^2 = 34
=> 2(x - 4) + 2(y - 5) y' = 0
=> y' = -(x - 4)/(y - 5)

at P(-1,2), y' = -(-1 -4)/(2 - 5)
= y' = -5/3 ......gradient of the tangent at P

(c) Using y - y1 = m(x - x1)
=> y - 2 = -(5/3)(x + 1)
=> y - 2 = -(5/3)x - 5/3
=> (5/3)x + y - 1/3 = 0 ...........equation of the tangent line at P

5. P(2,2), Q(1,1), R(7,-3)

the perpendicular bisector of a line connecting two points is a line passing throught the midpoint of the line connecting two points perpendicular to the line.

(a) slope of PQ = (1 - 2)/(1 - 2) = 1
=> slope of perpendicular bisector = -1

midpoint(PQ) = ((x2 + x1)/2, (y2 + y1)/2) = ( 3/2 , 3/2)

so equation of perpendicular bisector of PQ:

y - y1 = m(x - x1)
=> y - 3/2 = -1(x - 3/2)
=> y - 3/2 = -x + 3/2
=> y = -x + 3

slope of PR = (-3 -2)/(7 - 2) = -1
=> slope of perpendicular bisector = 1

midpoint(PR) = ((7 + 2)/2 , (-3 + 2)/2) = ( 9/2 , -1/2 )

equation of bisector:
y - y1 = m(x - x1)
=> y + 1/2 = x - 9/2
=> y = x - 5

b) The center (I believe) will be where these lines intersect:

so we want:

x - 5 = -x + 3
=> 2x = 8
=> x = 4

when x = 4

y = 4 - 5 = -1

so center of circle is -1

Can anyone help me with these with step by step answers.
6. Note that C is on the x-axis and D is on the y-axis

Using (x + 3)^2 + (y - 3)^2 = 90

for x-intercept, y = 0
=> (x + 3)^2 + (0 - 3)^2 = 90
=> x^2 + 6x + 9 + 9 - 90 = 0
=> x^2 + 6x - 72 = 0
=> (x + 12)(x - 6) = 0
=> x = -12 or x = 6
since we are concerned with the positive region of the x-axis, x = 6
so C(6,0) so c = 6

for y-intercept, x = 0
=> (0 + 3)^2 + (y - 3)^2 = 90
=> 9 + y^2 - 6y + 9 - 90 = 0
=> y^2 - 6y - 72 = 0
so y = 12 or y = -6
we want the positive region of the y-axis, so y = 12
D(0,12) so d = 12

7. x^2 + y^2 = 80
=> y = sqrt(80 - x^2) we show that the line y = 30 - 2x does not meet this circle.

set the y's equal, we get:

30 - 2x = sqrt(80 - x^2)
=> (30 - 2x)^2 = 80 - x^2
=> 900 - 120x + 4x^2 = 80 - x^2
=> 5x^2 - 120x + 820 = 0
x = [120 +/- sqrt(14400 - 16400)]/10
=> x = [120 +/- sqrt(-2000)]/10

I don't feel like simplifying any further, i'm tired. just note that there is a negative number under the square root, so the two graphs do not meet at any real number coordinates and therefore do not meet.

I hope you're learning something from my explanations.
. . I really hate doing someone's homework.

4) The point P(-1,2) lies on the circle with center C(4,5).
(a) Find the equation of the circle.
We already have: .(x - 4)² + (y - 5)² .= .

All we need is the radius.
It's the distance between P and C.
I'll let you find it . . .

(b) Find the gradient of the tangent at P.
This is a bit more subtle.

The tangent to a circle is perpendicular to the radius at that point.

The radius is CP, and its slope is: .(5 - 2)/(4 + 1) .= .3/5

Therefore, the slope (gradient) of the tangent is: .-5/3

(c) Find the equation of the tangent at P in the form: ax + by + c = 0
We know point P(-1,2) and the slope, m = -5/3

Use the Point-Slope Formula to write the equation of that line:
. . y - 2 .= .(-5/3)(x + 1) . . 5x + 3y - 1 .= .0

5) The points P(2,2), Q(1,1) and R(7,-3) lies on a circle.
(a) Find the equation of the perpendicular bisector of: (i) PQ .(ii) PR
(i) The midpoint of PQ is: .x = ½(2 + 1) = 3/2, . y = ½(2 + 1) = 3/2
The slope of PQ is: .(2 - 1)/(2 - 1) .= .+1
The perpendicular slope is: .m = -1
. . The equation is: .y - 3/2 .= .-1(x - 3/2) . . y .= .-x + 3

(ii) The midpoint of PR is: .x = ½(2 + 7) = 9/2, . y = ½(2 - 3) = -1/2

The slope of PR is: .(-3 -2)/(7 - 2) .= .-1
The perpendicular slope is: .m = +1
The equation is: .y + 1/2 .= .1(x - 9/2) . . y .= .x - 5

(b) Hence, find the coordinates of the center of the circle.
The center is at the intersection of the two perpendicular bisectors.

Solve the system of equations: .y .= .-x + 3, . y .= .x - 5

I'll let you do that . . .

6) The circle (x + 3)² + (y - 3)² .= .90
meets the positive coordinates axes at C(c,0) and d(0,d).

(a) Find the values of C and D.
For an x-intercept, let y = 0.
. . (x + 3)² + (0 - 3)² .= .90 . . (x + 3)² + 9 .= .90
. . (x + 3)² .= .81 . . x + 3 .= .±9 . . x .= .6, -12
Therefore: .C(6,0)

For a y-intecept, let x = 0.
. . (0 + 3)² + (y - 3)² .= .90 . . 9 + (y - 3)² .= .90
. . (y - 3)² .= .81 . . y - 3 .= .±9 . . y .= .12, -6
Therefore: .D(0,12)

(b) Find the area of ∆OCD, where O is the origin.
∆OCD is a right triangle with base 6 and height 12.

You finish it . . .