given circle and line equations, find point closest to K

P

Peter Burnes

New member
Joined
Jun 20, 2006
Messages
15
I would like a bit of help understanding how to do this question please:

. . .Given the circle C: x^2 + y^2 + 4x + 4y - 17 = 0 and the line Z: 4x + 3y = 12,
. . .find the co-ordinates of the point on S that is closest to K.

I've showed that Z and C don't intersect by calculating the center of C (-2, -2) the radius of C which is equal to 5 and then used the perpendicular formula to show that the distance from the centre to the line isn't the same as the radius.

I'm not sure how to go about getting the point that is closest to K. If they were intersecting then I could use simultaneous equations.

Would this method work? If I got the equation of a perpendicular line to Z and used simultaneous equations with that and the equation of the circle? What point on the line would I use to get the equation of the line perpendicular to Z? Any thoughts?

Thank you!
 
Re: Any help with the equation of a circle and a line questi

Peter Burnes said:
Given the circle C: x^2+y^2+4x+4y-17 =0 and the line Z: 4x+3y=12. I'm asked to find the co-ordinates of the point on S that is closest to K
Z and C don't intersect by calculating the center of C (-2,-2) the radius of C which is equal to 5 and then used the perpendicular formula to show that the distance from the centre to the line is greater than the radius.

Get a perpendicular line to Z through the center of the circleand used simultaneous equations with that and the equation of the circle?
Click to expand...
 
So I should get the equation of a line through the center of the circle with inverse slope to the line and then use simultaneous equations with the equation of the circle? correct?
 
Peter Burnes said:
So i should get the equation of a line through the center of the circle with inverse slope to the line and then use simultaneous equations with the equation of the circle? correct?
Click to expand...
Correct.
 
I got the new line 3x-4y-2=0

I substituted x with (4y+2)/3 into my equation for a circle.

Multiplying ot I get 25y^2+68y+11=0

The answer to the question is supposed to be the point (2,1).

What do you think? Am I going about it the wrong way or could the answer be wrong?

Thanking you!
 
When I solved the two equations, I got y=-5 or y=1.
 
Peter Burnes said:
3x-4y-2=0 Did you get this as the line you worked from?
Click to expand...
Yes indeed.
SOLVE:\(\displaystyle \left( {\frac{{4y + 2}}{3}} \right)^2 + y^2 + 4\left( {\frac{{4y + 2}}{3}} \right) + 4y - 17 = 0\)
 
Hey pka!
Well i managed to solve the equation and I got the points (2,1) and (-6,-5)
It's great when they work out alright!
Thanks very much for your guidence! Very much appreciated.
I was wondering one last thing.
Is there any special test to do to decide which of these points is the one that is closet to the line? I say this because I solved this question without been given a diagram.
Although I know the answer is 2, 1 ... how would I know this if I wasn't told the answer?

Thanks again pka
 
Peter Burnes said:
the points (2,1) and (-6,-5)
Although I know the answer is 2, 1 ... how would I know this if I wasn't told the answer?
Click to expand...
Given the point P and line l the distance from P to l is
\(\displaystyle \L
\begin{array}{l}
P(r,s)\quad l:Ax + By + C = 0 \\
D(P,l) = \frac{{\left| {Ar + Bs + C} \right|}}{{\sqrt {A^2 + B^2 } }} \\
\end{array}\).
 
You must log in or register to reply here.
Top