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Magazine Chapter 1: Problem 23
Find an equation of the line which is tangent to the circle \(x^{2}+y^{2}-4 x+6 y-12=0\) at the point \((5,1)\).
Short Answer
Expert verified
The equation of the tangent line is \[3x + 4y = 19\].
Step by step solution
01
Rewrite the Circle Equation
The given circle equation is: \[x^{2}+y^{2}-4 x+6 y-12=0\]To rewrite it in standard form, complete the square for the x and y terms.
02
Complete the Square
Complete the square for the x-terms and y-terms separately:\[x^{2} - 4x \rightarrow (x - 2)^{2} - 4\]\[y^{2} + 6y \rightarrow (y + 3)^{2} - 9\]The equation becomes:\[(x - 2)^{2} - 4 + (y + 3)^{2} - 9 - 12 = 0\]Simplify it to get: \[(x - 2)^{2} + (y + 3)^{2} = 25\]
03
Identify the Circle Center and Radius
From the equation \[(x - 2)^{2} + (y + 3)^{2} = 25\]Identify the center and radius:Center: \((2, -3)\)Radius: \(5\) (since \(25 = 5^{2}\))
04
Find the Slope of the Radius
The slope of the radius from the center \((2, -3)\) to the point \((5, 1)\) is given by:\[\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-3)}{5 - 2} = \frac{4}{3}\]
05
Determine the Slope of the Tangent Line
The slope of the tangent line is the negative reciprocal of the slope of the radius. Therefore, the tangent slope is:\[\text{slope}_{\text{tangent}} = -\frac{3}{4}\]
06
Form the Tangent Line Equation
Use the point-slope form \(y - y_1 = m (x - x_1)\) where \((x_1, y_1)\) is \((5, 1)\) and \(m\) is \(-\frac{3}{4}\):\[y - 1 = -\frac{3}{4} (x - 5)\]Simplify this to find the equation of the tangent line:\[4(y - 1) = -3(x - 5)\]\[4y - 4 = -3x + 15\]\[3x + 4y = 19\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
equation of circle
An equation of a circle in algebra expresses the set of all points at a fixed distance (radius) from a center point. The general form is given by:
\[x^2 + y^2 + Dx + Ey + F = 0\]
To convert this general form into the standard form \[(x - h)^2 + (y - k)^2 = r^2\], we need to complete the square. The terms \((x - h)\) and \((y - k)\) represent translations of the circle along the x and y axes, where (h, k) is the center. Here鈥檚 how that fits in:
\[x^2 + y^2 - 4x + 6y - 12 = 0\]
to the standard form
\[(x - 2)^2 + (y + 3)^2 = 25\]
which shows the circle鈥檚 center at (2, -3) and a radius of 5.
\[x^2 + y^2 + Dx + Ey + F = 0\]
To convert this general form into the standard form \[(x - h)^2 + (y - k)^2 = r^2\], we need to complete the square. The terms \((x - h)\) and \((y - k)\) represent translations of the circle along the x and y axes, where (h, k) is the center. Here鈥檚 how that fits in:
- Rewrite the x-terms and y-terms to help identify the circle鈥檚 properties.
- Check for middle terms of x and y to determine the need for completing the square.
\[x^2 + y^2 - 4x + 6y - 12 = 0\]
to the standard form
\[(x - 2)^2 + (y + 3)^2 = 25\]
which shows the circle鈥檚 center at (2, -3) and a radius of 5.
slope calculation
The slope is a measure of the steepness and direction of a line. It鈥檚 crucial for finding the tangent line to a circle at a specific point. Here's how to find the slope, given two points:
\[\text{slope} = \frac{1 - (-3)}{5 - 2} = \frac{4}{3}\]
This tells us how steep the radius is. To find the tangent line, remember:
\[\text{slope}_{\text{tangent}} = -\frac{3}{4}\].
- Identify your two points as \((x_1, y_1)\) and \((x_2, y_2)\).
- Use the formula: \(\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}\).
- In our example, we calculated the slope of the radius.
\[\text{slope} = \frac{1 - (-3)}{5 - 2} = \frac{4}{3}\]
This tells us how steep the radius is. To find the tangent line, remember:
- Tangent lines are perpendicular to the radius at the point of tangency.
\[\text{slope}_{\text{tangent}} = -\frac{3}{4}\].
point-slope form
The point-slope form of a line鈥檚 equation makes it easy to write the line鈥檚 equation when you know one point on the line and its slope. The formula is:
\[y - y_1 = m(x - x_1)\]
where:
\[y - 1 = -\frac{3}{4}(x - 5)\].
After simplifying, the equation of the tangent line becomes:
\[4(y - 1) = -3(x - 5)\]
which further simplifies to:
\[3x + 4y = 19\].
This equation represents the tangent line to the circle at the given point.
\[y - y_1 = m(x - x_1)\]
where:
- \((x_1, y_1)\) is a known point
- \(m\) is the slope
\[y - 1 = -\frac{3}{4}(x - 5)\].
After simplifying, the equation of the tangent line becomes:
\[4(y - 1) = -3(x - 5)\]
which further simplifies to:
\[3x + 4y = 19\].
This equation represents the tangent line to the circle at the given point.
completing the square
Completing the square is a method used to transform quadratic equations into a standard form that represents parabolas, circles, ellipses, or hyperbolas. For circles, it helps us reveal the circle鈥檚 center and radius in the equation. Follow these steps to complete the square:
\[x^2 + y^2 - 4x + 6y - 12 = 0\].
We completed the square for each variable separately:
\[x^2 - 4x -> (x - 2)^2 - 4\],
\[y^2 + 6y -> (y + 3)^2 - 9\].
Inserting these back into the equation, we get:
\[(x - 2)^2 - 4 + (y + 3)^2 - 9 = 12\].
Simplifying, we obtain:
\[(x - 2)^2 + (y + 3)^2 = 25\].
This reveals the circle鈥檚 center at (2, -3) and radius 5.
- Group x and y terms together.
- Add and subtract the necessary constants to make perfect square trinomials.
- Simplify any remaining terms.
\[x^2 + y^2 - 4x + 6y - 12 = 0\].
We completed the square for each variable separately:
\[x^2 - 4x -> (x - 2)^2 - 4\],
\[y^2 + 6y -> (y + 3)^2 - 9\].
Inserting these back into the equation, we get:
\[(x - 2)^2 - 4 + (y + 3)^2 - 9 = 12\].
Simplifying, we obtain:
\[(x - 2)^2 + (y + 3)^2 = 25\].
This reveals the circle鈥檚 center at (2, -3) and radius 5.
