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Sagot :
To find the equation of the line tangent to the circle [tex]\(x^2 + y^2 = 169\)[/tex] at the point [tex]\((5, 12)\)[/tex], follow these steps:
1. Identify the Circle's Characteristics: The given equation [tex]\(x^2 + y^2 = 169\)[/tex] represents a circle with its center at the origin [tex]\((0, 0)\)[/tex] and radius [tex]\(r\)[/tex]. Since [tex]\(169\)[/tex] is the square of [tex]\(13\)[/tex], the radius [tex]\(r\)[/tex] is [tex]\(13\)[/tex].
2. Determine the Slope of the Radius: The radius joins the center of the circle [tex]\((0, 0)\)[/tex] to the given point [tex]\((5, 12)\)[/tex]. The slope of this radius is given by the formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ \text{slope of the radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 0}{5 - 0} = \frac{12}{5} \][/tex]
3. Calculate the Slope of the Tangent Line: The tangent line at any point on a circle is perpendicular to the radius at that point. Perpendicular slopes are negative reciprocals of each other. Therefore, the slope of the tangent line [tex]\(m_{\text{tangent}}\)[/tex] is:
[tex]\[ m_{\text{tangent}} = -\frac{1}{\frac{12}{5}} = -\frac{5}{12} \][/tex]
4. Form the Equation of the Tangent Line: The standard form for the equation of a line is [tex]\(y = mx + b\)[/tex]. Here, we know the slope [tex]\(m = -\frac{5}{12}\)[/tex] and the point [tex]\((5, 12)\)[/tex] through which the line passes. Substitute [tex]\(x = 5\)[/tex], [tex]\(y = 12\)[/tex], and [tex]\(m = -\frac{5}{12}\)[/tex] into the equation to solve for the y-intercept [tex]\(b\)[/tex]:
[tex]\[ 12 = -\frac{5}{12} \cdot 5 + b \][/tex]
5. Solve for the y-intercept [tex]\(b\)[/tex]:
[tex]\[ 12 = -\frac{25}{12} + b \][/tex]
To isolate [tex]\(b\)[/tex], add [tex]\(\frac{25}{12}\)[/tex] to both sides:
[tex]\[ 12 + \frac{25}{12} = b \][/tex]
Convert [tex]\(12\)[/tex] to a fraction with a common denominator of [tex]\(12\)[/tex]:
[tex]\[ 12 = \frac{144}{12} \][/tex]
Adding the fractions:
[tex]\[ \frac{144}{12} + \frac{25}{12} = \frac{169}{12} \][/tex]
Therefore, the y-intercept [tex]\(b\)[/tex] is:
[tex]\[ b = \frac{169}{12} \approx 14.0833 \][/tex]
Putting it all together, the equation of the tangent line in slope-intercept form is:
[tex]\[ y = -\frac{5}{12}x + \frac{169}{12} \][/tex]
The final equation of the tangent line to the circle at the point [tex]\((5, 12)\)[/tex] is:
[tex]\[ y = -0.4167x + 14.0833 \][/tex]
Hence, the slope of the tangent line is approximately [tex]\(-0.4167\)[/tex] and the y-intercept is approximately [tex]\(14.0833\)[/tex].
1. Identify the Circle's Characteristics: The given equation [tex]\(x^2 + y^2 = 169\)[/tex] represents a circle with its center at the origin [tex]\((0, 0)\)[/tex] and radius [tex]\(r\)[/tex]. Since [tex]\(169\)[/tex] is the square of [tex]\(13\)[/tex], the radius [tex]\(r\)[/tex] is [tex]\(13\)[/tex].
2. Determine the Slope of the Radius: The radius joins the center of the circle [tex]\((0, 0)\)[/tex] to the given point [tex]\((5, 12)\)[/tex]. The slope of this radius is given by the formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ \text{slope of the radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 0}{5 - 0} = \frac{12}{5} \][/tex]
3. Calculate the Slope of the Tangent Line: The tangent line at any point on a circle is perpendicular to the radius at that point. Perpendicular slopes are negative reciprocals of each other. Therefore, the slope of the tangent line [tex]\(m_{\text{tangent}}\)[/tex] is:
[tex]\[ m_{\text{tangent}} = -\frac{1}{\frac{12}{5}} = -\frac{5}{12} \][/tex]
4. Form the Equation of the Tangent Line: The standard form for the equation of a line is [tex]\(y = mx + b\)[/tex]. Here, we know the slope [tex]\(m = -\frac{5}{12}\)[/tex] and the point [tex]\((5, 12)\)[/tex] through which the line passes. Substitute [tex]\(x = 5\)[/tex], [tex]\(y = 12\)[/tex], and [tex]\(m = -\frac{5}{12}\)[/tex] into the equation to solve for the y-intercept [tex]\(b\)[/tex]:
[tex]\[ 12 = -\frac{5}{12} \cdot 5 + b \][/tex]
5. Solve for the y-intercept [tex]\(b\)[/tex]:
[tex]\[ 12 = -\frac{25}{12} + b \][/tex]
To isolate [tex]\(b\)[/tex], add [tex]\(\frac{25}{12}\)[/tex] to both sides:
[tex]\[ 12 + \frac{25}{12} = b \][/tex]
Convert [tex]\(12\)[/tex] to a fraction with a common denominator of [tex]\(12\)[/tex]:
[tex]\[ 12 = \frac{144}{12} \][/tex]
Adding the fractions:
[tex]\[ \frac{144}{12} + \frac{25}{12} = \frac{169}{12} \][/tex]
Therefore, the y-intercept [tex]\(b\)[/tex] is:
[tex]\[ b = \frac{169}{12} \approx 14.0833 \][/tex]
Putting it all together, the equation of the tangent line in slope-intercept form is:
[tex]\[ y = -\frac{5}{12}x + \frac{169}{12} \][/tex]
The final equation of the tangent line to the circle at the point [tex]\((5, 12)\)[/tex] is:
[tex]\[ y = -0.4167x + 14.0833 \][/tex]
Hence, the slope of the tangent line is approximately [tex]\(-0.4167\)[/tex] and the y-intercept is approximately [tex]\(14.0833\)[/tex].
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