do with fulllll fulll calculations Find the equations of the two tangents to the circle x2+y2+4x-6y+4=0 if the tangents pass throug the origin.

The given equation of the circle is x² + y² + 4x - 6y + 4 = 0.

We can rewrite this in the standard form of a circle equation (x - h)² + (y - k)² = r², where (h, k) is the center of the circle and r is the radius.

Comparing the given equation with the standard form, we get:

(x + 2)² + (y - 3)² = r²

This simplifies to:

x² + 4x + 4 + y² - 6y + 9 = r²

Comparing this with the given equation, we find that r² = 9, so r = 3.

The center of the circle is at (-2, 3).

The equation of a line passing through the origin and a point (x1, y1) is y = (y1/x1)x.

Since the tangents pass through the origin, the slopes of the tangents are given by m1 and m2, where m1*m2 = -1 (because the tangents are perpendicular to the radius at the point of tangency).

Let's denote the point of tangency as (x1, y1). The distance from the center of the circle to the point of tangency is equal to the radius, so we have:

(x1 + 2)² + (y1 - 3)² = 9

This simplifies to:

x1² + 4x1 + 4 + y1² - 6y1 + 9 - 9 = 0

x1² + 4x1 + y1² - 6y1 + 4 = 0

The equation of the line passing through the origin and the point of tangency is y = mx, so y1 = m*x1.

Substituting y1 = m*x1 into the equation above, we get:

x1² + 4x1 + (mx1)² - 6mx1 + 4 = 0

This is a quadratic equation in x1, so its discriminant must be zero (since there is only one point of tangency).

The discriminant is (4 - 6m)² - 4(1 + m²) = 0.

Solving this equation for m gives the slopes of the tangents, and substituting these slopes into the equation y = mx gives the equations of the tangents.