do it with full full fulllllllcalculations.Consider the cirele x2+y2+x+7y+4-0. (a) Show that k(-3, -2) is a point on the circle (b) Find the equation of the tangent to the cirele at K

(a) To show that the point K(-3, -2) lies on the circle with equation x² + y² + x + 7y + 4 = 0, we substitute the x and y coordinates of point K into the equation of the circle. If the left side equals the right side after substitution, then the point lies on the circle.

Substituting x = -3 and y = -2 into the equation:

(-3)² + (-2)² + (-3) + 7*(-2) + 4 = 0 9 + 4 - 3 - 14 + 4 = 0 0 = 0

Since the left side equals the right side, the point K(-3, -2) lies on the circle.

(b) The 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 equation of the circle can be rewritten as x² + y² + x + 7y + 4 = 0 as (x + 1/2)² + (y + 7/2)² = r². Comparing the two equations, we find that the center of the circle is (-1/2, -7/2).

The slope of the radius from the center of the circle to the point K is given by (y2 - y1) / (x2 - x1) = ((-2) - (-7/2)) / ((-3) - (-1/2)) = -3.

The tangent line at a point on the circle is perpendicular to the radius at that point. Therefore, the slope of the tangent line is the negative reciprocal of the slope of the radius, which is 1/3.

The equation of the tangent line can be found using the point-slope form of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Substituting m = 1/3 and the coordinates of point K into this equation gives the equation of the tangent line:

y - (-2) = 1/3(x - (-3)) y + 2 = 1/3x + 1 y = 1/3x - 1

So, the equation of the tangent to the circle at K is y = 1/3x - 1.