A CIRCLE HAS EQUATION X^2 +4X + 10Y -8 =0,FIND THE EQUATION OF THE TANGENT AT (3,1)
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The point (3,1) does not satisfy the equation.
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To find the equation of the tangent to the circle at the point (3,1), we need to first determine the slope of the tangent line.
Step 1: Convert the equation of the circle to general form
Given equation of the circle: x^2 + 4x + 10y - 8 = 0
To convert this to general form (Ax + By + C = 0), we can rearrange the terms:
x^2 + 4x + 10y - 8 = 0
x^2 + 4x - 10y + 8 = 0
Step 2: Identify the slope of the tangent line
The gradient of a tangent to the circle at a given point is equal to the negative reciprocal of the gradient of the radius (line connecting the center of the circle to the given point).
The given point is (3,1), so we need to find the gradient of the radius first.
The center of the circle can be found by completing the square for the x and y terms, as the general form: x^2 + 4x + 10y - 8 = 0 is equivalent to: (x^2 + 4x + 4) + 10y - 8 = 4
(x + 2)^2 + 10y - 12 = 4
(x + 2)^2 + 10y = 16
Comparing this equation to the standard form of a circle x^2 + y^2 = r^2, we can infer:
(x + 2)^2 + 10y = 16
r^2 = 16
r = sqrt(16) = 4
So, the center of the circle is (-2, 0).
To find the gradient of the radius at the given point, we can use the formula:
Gradient = (y2 - y1) / (x2 - x1)
Using the center of the circle (-2, 0) and the given point (3, 1):
Gradient = (1 - 0) / (3 - (-2))
Gradient = 1 / 5
Therefore, the gradient of the tangent line is the negative reciprocal of the gradient of the radius at the given point, which is -5.
Step 3: Determine the equation of the tangent line
Using the point-slope form of a linear equation (y - y1) = m(x - x1), where (x1, y1) is the given point and m is the slope:
(y - 1) = -5(x - 3)
Distributing, we get:
y - 1 = -5x + 15
Rearranging the equation in the slope-intercept form (y = mx + b):
y = -5x + 16
Therefore, the equation of the tangent line at the point (3, 1) is y = -5x + 16.
To find the equation of the tangent line to a circle at a given point, you need to follow these steps:
Step 1: Rewrite the equation of the circle in the standard form.
Given circle equation: x^2 + 4x + 10y - 8 = 0
To rewrite it in the standard form, complete the square for both x and y terms.
x^2 + 4x = -10y + 8
(x + 2)^2 - 4 = -10y + 8
(x + 2)^2 = -10y + 12
Step 2: Find the slope of the line tangent to the circle at the given point (3, 1).
The slope of the tangent line can be found by taking the derivative of the circle equation with respect to x and evaluating it at the given point (3, 1).
Differentiating the circle equation:
2(x + 2) = -10(dy/dx)
To find dy/dx, substitute x = 3 and y = 1:
2(3 + 2) = -10(dy/dx)
10 = -10(dy/dx)
(dy/dx) = -10/10
(dy/dx) = -1
So, the slope of the tangent line is -1.
Step 3: Use the point-slope form of a line to find the equation of the tangent line.
Using the formula:
y - y1 = m(x - x1)
Substituting the point (3, 1) and the slope m = -1:
y - 1 = -1(x - 3)
y - 1 = -x + 3
Rearranging the equation:
y + x = 4
Therefore, the equation of the tangent line to the circle at the point (3, 1) is y + x = 4.