the equation of a circle is x^2+y^2+kx-8y-(2k-3)=0 and its radius is k+1,wherek>0.find the value of k and hence the centre of the circle.show that the line segment joining P(-6,7)and q(2,1)is a diameter of the circle

The general equation of a circle is given by (x-h)² + (y-k)² = r², where (h,k) is the center of the circle and r is the radius.

The given equation of the circle is x² + y² + kx - 8y - (2k-3) = 0. We can rewrite this in the form of the general equation by completing the square.

First, group the x and y terms together:

(x² + kx) + (y² - 8y) = 2k - 3.

To complete the square, we need to add and subtract (k/2)² = k²/4 to the x terms and add and subtract (8/2)² = 16 to the y terms:

[(x² + kx + k²/4) - k²/4] + [(y² - 8y + 16) - 16] = 2k - 3.

This simplifies to:

[(x + k/2)² - k²/4] + [(y - 4)² - 16] = 2k - 3.

Combine the constants on the left side:

(x + k/2)² + (y - 4)² = k²/4 + 16 + 2k - 3.

This is now in the form of the general equation of a circle, so we can see that the center of the circle is (-k/2, 4) and the radius is √(k²/4 + 16 + 2k - 3).

Given that the radius is k + 1, we can set up the equation:

√(k²/4 + 16 + 2k - 3) = k + 1.

Squaring both sides to get rid of the square root gives:

k²/4 + 16 + 2k - 3 = k² + 2k + 1.

Solving this equation for k gives k = 4.

Substituting k = 4 into the center of the circle gives the center as (-2, 4).

The line segment joining P(-6,7) and Q(2,1) would be a diameter of the circle if the midpoint of PQ is the center of the circle. The midpoint of PQ is ((-6+2)/2, (7+1)/2) = (-2, 4), which is the center of the circle. Therefore, the line segment PQ is a diameter of the circle.