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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

Best Answer 

 #1
avatar+92781 
+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
 #1
avatar+92781 
+10
Best Answer

 

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
avatar+87301 
0

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

 

CPhill  Jan 27, 2015
 #3
avatar+92781 
0

Thanks Chris :)

Melody  Jan 27, 2015

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