Co-ordinate geometry: find radius of circle, and....

[davidg]

can any1 help me with this question! I can do part i but not part ii.

Circle C has centre(5,-1)
The line L:3x-4y=11=0 is a tangent to C
(i)Show that the radius of C is 6
(ii)The line x+py+1=0 is also a tangent to C. Find the two possible values of p

 

[pka]

Re: Co-ordinate geo

Circle C has centre(5,-1)
The line L:3x-4y=11=0 is a tangent to C
(i)Show that the radius of C is 6
(ii)The line x+py+1=0 is also a tangent to C. Find the two possible values of p

For (i). What is the distance of (5,-1) to L?

 

[galactus]

Re: Co-ordinate geo

(ii)The line x+py+1=0 is also a tangent to C. Find the two possible values of p

After you prove the radius is, indeed, 6 use the formula for the distance between a point and a line, set equal to 6 and solve for p.

Here's one of the p values graphed. Your mission is to find the equation of this line as well as the other.

In part i, I am going to take a leap of faith and assume your typo was 3x-4y+11=0.

 

[jacket81]

You made a typo on equation for line L.
3x-4y=11=0????

What exactly is it?

 

[davidg]

O my god how embarassing I ask ye for help and then i do that! Sorry the line L is 3x-4y+11=0
thanks for all the help

 

[jacket81]

Okay, its 3x-4y+11=0, or 3x+11=4y. SO, to find its slope, divide by 4 to get (3/4)x+(11/4) = y. So its slope is 3/4. So, a line perpendicualr to it is -4/3.
Put in y=mx+b form, with point (5,-1), -1=((-4/3)(5))+b, so b=(20/3)-1=17/3.
So, the line is y=(-4/3)x+(17/3).
Now maybe multiply both sides by 3 to get 3y=-4x+17.
Now find where this equation and the other equation, 3x+11=4y, intersect.
They intersect at x=7/5, and y=19/5.
Now the distance between this point and (5,-1) should be 6.
Squareroot((5-(7/5)^2+(-1-(19/5)0^2) = sqr((18/5)^2+(-24/5)^2) = 6.

 

[pka]

Now the distance between this point and (5,-1) should be 6.

Isn’t it a lot easier to just note that \(\displaystyle \frac{{\left| {(3)(5) + ( - 4)( - 1) + 11} \right|}}{{\sqrt {3^2 + 4^2 } }} = 6\) ?

Using the distance formula for a point and line.

 

[davidg]

Thanks for all your help. The way i went about it was that i used the perpendicular distance formula. So used the centre point and a=1 b=p and c=1 and the radius of 6. and it just wont solve out to give me two values for p ?

Any more advice? :wink: