cartesian inequality

**samdmansam** · May 15th 2008, 06:17 AM

Find the Cartesian inequation for the region represented by
abs(z+9-9i)<=1/7*abs(z-3-2i)

Please put your answer in a "natural" form
What is the centre of that circle for real numbers x, y
What is the radius of the circle

The locus of points satisfied by the relation is

If any one knows where there are good online math resources to do with cartesian inequations can you let me know
Thanks

**Soroban** · May 15th 2008, 07:40 AM

Hello, samdmansam!

Find the Cartesian inequality for the region represented by:
$|z+9-9i| \;\leq \; \frac{1}{7}|z-3-2i|$

Write your answer in a "natural" form.
What is the centre of that circle?
What is the radius of the circle?

We have: . $\bigg|z-(-9+9i)\bigg| \;\leq\; \frac{1}{7}\bigg|z - (3+2i)\bigg|$

We are given two points on the complex plane: . $A(-9,9)\:\text{ and }\:B(3,2)$

We want all points $P(x,y)$ such that: . $d(PA) \:\leq \:\frac{1}{7}\,d(PB)$

. . $d(PA) \;=\;\sqrt{(x+9)^2 + (y-9)^2}\qquad d(PB) \;=\;\sqrt{(x-3)^2 + (y-2)^2}$

We have: . $\sqrt{(x+9)^2 + (y-9)^2} \;\leq \;\frac{1}{7}\sqrt{(x-3)^2 + (y-2)^2}$

. . which simplifies to: . $48x^2 + 888x + 48y^2 - 878y + 7925 \;\leq \;0$

Now complete the square to find the center and radius of this circle.
. . We want all points on and inside the circle.

Good luck!

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