do these with full calculation steps,show answers.Two tangents are drawn to the circlex2+y2+2x+4y-12 = 0. If the two tangents are perpendicular tothe straight line x-4y-7= 0, find (a) the equations of the two tangents, (b) the coordinates of the point of contact for each tangent found in (a)

First, let's rewrite the equation of the circle in standard form. The equation of the circle is given as:

x² + y² + 2x + 4y - 12 = 0

We can rewrite this as:

(x² + 2x + 1) + (y² + 4y + 4) - 12 - 1 - 4 = 0

This simplifies to:

(x + 1)² + (y + 2)² = 17

So, the center of the circle is at (-1, -2) and the radius is √17.

The equation of the line is given as x - 4y - 7 = 0. The slope of this line is 1/4. Since the tangents are perpendicular to this line, the slopes of the tangents are the negative reciprocal of 1/4, which is -4.

The equation of a line with slope m that passes through a point (h, k) is given by y - k = m(x - h). So, the equations of the tangents are:

y + 2 = -4(x + 1) and y + 2 = -4(x + 1)

Solving these equations for y gives:

y = -4x - 6 and y = -4x - 6

These are the equations of the two tangents.

To find the points of contact, we use the fact that the tangent line is perpendicular to the radius at the point of contact. The equation of the line through two points (x1, y1) and (x2, y2) is given by (y - y1) = m(x - x1), where m is the slope of the line. So, the equations of the lines through the center of the circle and the points of contact are:

y + 2 = -1/4(x + 1) and y + 2 = -1/4(x + 1)

Solving these equations simultaneously with the equations of the tangents gives the points of contact:

(-3, -10) and (-3, -10)