# Maths Homework!Watch

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#1

1)find the least and greatest distances of the origin from the circumference of the cirle with equation (x-12)^2 + (y-5)^2 = 25

2)the point P (x,y) moves so that its distance from the point (3,4) is twice its distance from the origin O. Find in cartesian form an equation for the path of P
0
15 years ago
#2
1) Circle has centre (12,5) and radius 5. The distance to the centre is sqrt(12^2+5^2)=13. The least distance is, therefore, 13-5=8 and the greatest is 13+5=18.

2) I won't write out the solution to this but basically you need to find an expression for the distance from the origin and an expression for the distance from the point (3,4) using pythagoras. Once you have two expressions, you set up an equation like so:

(distance from (3,4)) = 2(distance from the origin)

This will be your equation since it will be in x and y, rearranging it will give you the required equation.
0
15 years ago
#3
Picture for (1):
0
15 years ago
#4
(Original post by mikesgt2)
(distance from (3,4)) = 2(distance from the origin)
It's neater to use the equivalent form

[distance from (3, 4)]^2 = 4 [distance from the origin]^2

so we don't have to manipulate square roots. That form gives

(x - 3)^2 + (y - 4)^2 = 4 (x^2 + y^2)
3x^2 + 3y^2 + 6x + 8y - 25 = 0
3(x + 1)^2 + 3(y + 4/3)^2 = 100/3
(x + 1)^2 + (y + 4/3)^2 = (10/3)^2.

So we get the circle with centre (-1, -4/3) and radius 10/3.
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