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**+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**+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