The equations of the two tangents from(-5,-4) to the circle x2 y2 4x 6y 8=0?

Question

Answer

Equations of Tangents to a Circle

To find the equations of the two tangents from (-5,-4) to the circle x^2 + y^2 + 4x + 6y + 8 = 0, we need to follow these steps:

Step 1: Find the Center of the Circle

We can find the center of the circle by completing the square for both x and y terms:

x^2 + 4x + y^2 + 6y + 8 = 0

(x + 2)^2 - 4 + (y + 3)^2 - 9 + 8 = 0

(x + 2)^2 + (y + 3)^2 = 5

So, the center of the circle is (-2,-3) and its radius is sqrt(5).

Step 2: Find the Distance from the Point to the Center

The distance from (-5,-4) to the center (-2,-3) is:

sqrt[(x2 - x1)^2 + (y2 - y1)^2]

sqrt[(-5 + 2)^2 + (-4 + 3)^2]

sqrt(10)

Step 3: Find the Slope of the Line from the Point to the Center

The slope of the line from (-5,-4) to (-2,-3) is:

(y2 - y1)/(x2 - x1)

(-3 - (-4))/(-2 - (-5))

1/3

Step 4: Find the Slope of the Tangents

The slope of the tangents is perpendicular to the slope of the line from the point to the center. So, the slope of the tangents is:

-1/3

Step 5: Find the Equations of the Tangents

Using the point-slope form, the equations of the two tangents from (-5,-4) to the circle are:

y + 4 = (-1/3)(x + 5)

y + 4 = (-1/3)(x + 5) - 2sqrt(5)

Therefore, the equations of the two tangents from (-5,-4) to the circle x^2 + y^2 + 4x + 6y + 8 = 0 are y + 4 = (-1/3)(x + 5) and y + 4 = (-1/3)(x + 5) - 2sqrt(5).

Answer

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