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What are the equations of the tangent lines to circle x ^ 2 + y ^ 2 = 4x + 1 that are perpendicular to line 2x + y = 10​

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Answer:

y = (3a + sqrt(11a^2 - 8b + 41)) / 2

y = (3a - sqrt(11a^2 - 8b + 41)) / 2

Explanation:

To find the tangent lines that are perpendicular to the given line, we need to find the slope of the given line and then find the negative reciprocal of that slope. The negative reciprocal will be the slope of the tangent lines we are looking for.

The given line is 2x + y = 10, which can be rewritten as y = -2x + 10. Therefore, the slope of the given line is -2.

The center of the circle can be found by completing the square of the x and y terms. We have:

x^2 - 4x + y^2 = 1

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

Therefore, the center of the circle is (2,0), and the radius is sqrt(5).

Now, we can find the equation of the tangent lines. Let (a,b) be the point where the tangent line intersects the circle. Since the tangent line is perpendicular to the given line, its slope is the negative reciprocal of -2, which is 1/2. Therefore, the equation of the tangent line passing through (a,b) is:

y - b = (1/2)(x - a)

To find the points of intersection of this line and the circle, we substitute y = -2x + 10 into the equation of the tangent line and solve for x:

(x - a)^2 + (-2x + 10 - b)^2 = 5

Expanding and simplifying this equation, we get:

5x^2 - 4ax + 4a^2 + 4bx - 40b + 84 = 0

This is a quadratic equation in x, so we can solve for x using the quadratic formula:

x = [4a ± sqrt(16a^2 - 4(5)(4a^2 + 4b - 84))] / 10

Simplifying this expression, we get:

x = [2a ± sqrt(11a^2 - 8b + 41)] / 5

Now we can substitute these values of x into the equation of the tangent line to find the corresponding values of y:

y = -2x + 10

y = -a + b + (1/2)(2a ± sqrt(11a^2 - 8b + 41))

This gives us two equations for the tangent lines passing through (a,b), one for each value of x. We can simplify these equations to get:

y = (3a ± sqrt(11a^2 - 8b + 41)) / 2

Therefore, the equations of the tangent lines that are perpendicular to the given line are:

y = (3a + sqrt(11a^2 - 8b + 41)) / 2

y = (3a - sqrt(11a^2 - 8b + 41)) / 2

where (a,b) is any point on the circle (x - 2)^2 + y^2 = 5.

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