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`S1=x^2+y^2-2=0 ` and `S2=x^2+y^2+3x+3y-8=0` A straight line drawn through the point P...
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High School Teacher
To show that the midpoint of QR lies on the circle `x^2+y^2 +3/2x +3/2 y -5 =0` first find the co-ordinates of `S_(1)` and `S_(2)` at the center.
We know that `S_(1)` has the co-ordinates (0;0) from the equation :
`(x-h)^2 +(y-k)^2 = r^2` with the center at `(h;k)` and our
equation:`x^2+y^2=2` `(S_(1))` ;therfore h=0and k=0
From `S_(2)` we complete the square to find the center:
`therefore x^2+y^2+3x+3y-8=0` becomes
`x^2+3x +y^2+3y = 8`
To complete the square take the co efficient of x (ie 3) and the co-efficient of y (ie 3) and halve them `(3/2)` which result will appear in the brackets. As this is an equation we must do the same to the rught hand side by adding `3/2` . Note that it is squared because the `3/2` on the left hand side is also squared (in the bracket):
`(x+3/2)^2 + (y+3/2)^2 = 8 + (3/2)^2`
`therefore (x+3/2)^2 +(y+3/2)^2= 10 1/4`
In terms of the equation `(x-h)^2 +y-k)^2 = r^2` we now know that the center of the circle is at the co-ordinates `(-3/2; -3/2)`
Now to find the midpoint, we use the formula:
`M=((x_(1) +x_(2))/2); ((y_(1)+y_(2))/2)`
`therefore M=((0+(-3/2))/2); ` `((0+ (-3/2))/2)`
`therefore M=(-3/4; -3/4)`
To show that the midpoint lies on `x^2 +y^2+3/2x+3/2y-5=0`
complete the square to find the co-ordinates. Rearrange, halve `(3/2)` (ie = (`3/4)` for both x and y (in this instance) and square:
`therefore (x+3/4)^2 +(y+3/4)^2= 5+(3/4)^2`
`therefore (x+3/4)^2 +(y+3/4)^2 = 5 9/16`
We now have the center of the third equation - `(-3/4; -3/4)`
We know that this is ALSO the midpoint of QR and therefore, the line (QR) will cut the circle `x^2 +y^2 +3/2x+3/2y-5=0` and pass through the center.
Posted by durbanville on August 5, 2013 at 11:38 AM (Answer #1)
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