# Coordinate Geometry - circle question! One end of a diameter of a circle with equation `x^2 + y^2 - 4x + 4y = 2` is (3,1)...One end of a diameter of a circle with equation `x^2 + y^2 - 4x + 4y =...

Coordinate Geometry - circle question! One end of a diameter of a circle with equation `x^2 + y^2 - 4x + 4y = 2` is (3,1)...

One end of a diameter of a circle with equation `x^2 + y^2 - 4x + 4y = 2` is (3,1). Find the coordinates of the other end of the diameter.

I did a similar question that required to find the equation of the tangent to the circle with a equation given a point, which I firstly found the centre and found the slope of the tangent by making it negative reciprocal. I understand that type of question, but I am having trougble with this one. Help!

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`x^2 + y^2 - 4x + 4y = 2`

Express the equation in standard form `(x-h)^2 + (y-k)^2 = r^2` to be able to find the center of the circle.

`(x^2 - 4x) + (y^2 + 4y) = 2`

Perform completing the square method.

`(x^2 - 4x + 4) + (y^2 + 4y + 4) = 2 + 4 + 4`

`(x-2)(x-2) + (y+2)(y+2) = 10`

`(x-2)^2 + (y+2)^2 = 10`

So, center of the circle is (2,-2).

Note that center(h,k) is the midpoint of the diameter. To solve for the other end of the diameter, use the midpoint formula between two points.

`x_(mid) = (x_2+x_1)/2` and `y_(mid)= (y_2+y_1)/2`

Substitute center(2,-2) for (`x_(mid)` , `y_(mid)` ) and (3,1) for (`x_1` ,`y_1` ).

`2 = (x_2+3)/2` `-2 = (y_2+1)/2`

`4 = x_2 + 3 ` `-4 = y_2 + 1`

`1 = x_2 ` `-5 = y_2`

**Hence, the other end of the diameter is (1,-5).**