`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 cuts S1=0 and S2=0 again at the points Q and R respectively.

Show that the mid-point of QR lies on the circle

`x^2+y^2+ 3/2x+ 3/2y-5=0.`

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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.**

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