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# the equation of C is x^2+y^2-8x-20y+100=0 with cetre(4,10)

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the equation of C is x^2+y^2-8x-20y+100=0 with cetre(4,10)

given that  the slope and the x-intercept of a straight B are 1 and K respectively .If L cuts C at P and Q , find the ccoordinates of the mid -point of PQ in terms of K

Guest Jan 27, 2015

#1
+93683
+10

The equation of C is   $$x^2+y^2-8x-20y+100=0$$    with cetre(4,10)

given that  the slope and the x-intercept of a straight B are 1 and K respectively .If L cuts C at P and Q , find the ccoordinates of the mid -point of PQ in terms of K

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I am going to use the fact that the radius of a circle that bisects a chord is prependicular to the chord.

now the equation of the line is    y=1x+K

-------------------------------------------------------

So I want the equation of the line prependicular to  y=1x+k   through  (4,10)

the gradient will be -1

$$\\-1=\frac{y-10}{x-4}\\\\ -1(x-4)=y-10\\ -x+4=y-10\\ y=-x+4+10\\ y=-x+14$$

So now I want to find the intersection of  y=-x+14 and  y=x+k

$$\\-x+14=x+K\\ -2x=K-14\\ x=\frac{14-K}{2}\\\\ y=\frac{14-K}{2}+K\\\\ y=\frac{14-K}{2}+\frac{2K}{2}\\\\ y=\frac{14+K}{2}\\\\ \left(\frac{14-K}{2},\;\frac{14+K}{2}\right)$$

check

$$\\-x+14=\frac{K-14}{2}+\frac{28}{2}=\frac{K+14}{2}\\$$      Great it seems to work.

I went all around the world with this question before I got to the point.

It really does pay to think about problem questions before you barge head long into them.  LOL

Melody  Jan 27, 2015
#1
+93683
+10

The equation of C is   $$x^2+y^2-8x-20y+100=0$$    with cetre(4,10)

given that  the slope and the x-intercept of a straight B are 1 and K respectively .If L cuts C at P and Q , find the ccoordinates of the mid -point of PQ in terms of K

--------------------------------------------------------------------------

I am going to use the fact that the radius of a circle that bisects a chord is prependicular to the chord.

now the equation of the line is    y=1x+K

-------------------------------------------------------

So I want the equation of the line prependicular to  y=1x+k   through  (4,10)

the gradient will be -1

$$\\-1=\frac{y-10}{x-4}\\\\ -1(x-4)=y-10\\ -x+4=y-10\\ y=-x+4+10\\ y=-x+14$$

So now I want to find the intersection of  y=-x+14 and  y=x+k

$$\\-x+14=x+K\\ -2x=K-14\\ x=\frac{14-K}{2}\\\\ y=\frac{14-K}{2}+K\\\\ y=\frac{14-K}{2}+\frac{2K}{2}\\\\ y=\frac{14+K}{2}\\\\ \left(\frac{14-K}{2},\;\frac{14+K}{2}\right)$$

check

$$\\-x+14=\frac{K-14}{2}+\frac{28}{2}=\frac{K+14}{2}\\$$      Great it seems to work.

I went all around the world with this question before I got to the point.

It really does pay to think about problem questions before you barge head long into them.  LOL

Melody  Jan 27, 2015
#2
+90023
0

Very nice, Melody.....this one is a definite "Daily Wrap" candidate....!!!!

CPhill  Jan 27, 2015
#3
+93683
0

Thanks Chris :)

Melody  Jan 27, 2015