if a diameter of the circle 2x^2 +2y^2+4x+ky-27=0 passes through the point5,5 and the slope of the diameter is 0.5 find the value of k 回答問題

The equation of the circle is given by 2x^2 + 2y^2 + 4x + ky - 27 = 0.

First, we can rewrite this equation in the standard form of a circle equation, which is (x - h)^2 + (y - k)^2 = r^2.

Dividing the equation by 2, we get x^2 + y^2 + 2x + (k/2)y - 13.5 = 0.

Completing the square, we get (x + 1)^2 + (y + k/4)^2 = 14.25.

So, the center of the circle is at (-1, -k/4).

The diameter of the circle passes through the center of the circle and the point (5,5).

Since the slope of the diameter is 0.5, we can use the slope formula (y2 - y1) / (x2 - x1) = m to find the value of k.

Substituting the given values, we get (5 - (-k/4)) / (5 - (-1)) = 0.5.

Solving this equation for k, we get k = 20.

So, the value of k is 20.