Question

prove that the line 4x+y-5 = 0 is a tangent to the circle x^2+y^2+2x-y-3 = 0. Also find the point of contact.​

Answer 1

To prove that a line is a tangent to a circle, we need to find the point of intersection between the line and the circle and show that only one such point exists.

First, we'll solve for the point of intersection between the line and the circle. We'll substitute the equation for the line into the equation for the circle:

$4x + y - 5 = 0$

$x^2 + y^2 + 2x - y - 3 = 0$

Multiplying the first equation by 2 and subtracting the second equation, we get:

$2x^2 + 2y^2 + 4x - 3y = 2$

Completing the square for both x and y terms, we get:

$(2x + 1)^2 + (2y - 3)^2 = 4$

Thus, the point of intersection between the line and the circle is

(x, y) = (-0.5, 2).

To show that the line is a tangent to the circle, we need to find the slope of the line at the point of intersection and compare it to the slope of the tangent at that point on the circle.

The slope of the line is -4, and the slope of the tangent at the point of intersection on the circle can be found using the gradient at that point, which is the derivative of the equation of the circle with respect to x and y:

$dy/dx = -1 - 4x/2y + 2 = -2 + 2x/y$

Evaluating the gradient at

$(x, y) = (-0.5, 2)$,

we get:

$dy/dx = -2 + 2(-0.5)/2 = -2$

Since the slope of the line is equal to the gradient of the tangent at the point of intersection, we have shown that the line is indeed a tangent to the circle.

The point of contact is (x, y) = (-0.5, 2).

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