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Chapter 13: Problem 40

Find the equation of the circle that is tangent to the line \(3 x-4 y=-26\) at the point \((-2,5)\) and passes through the point \((5,-2) . \quad x^{2}+y^{2}-2 x-2 y-23=0\)

Short Answer

Expert verified
The circle with center \((-1, 2)\) is tangent and passes through \((5, -2)\).

Step by step solution

01

Identify the Center of the Circle

The given equation can be rewritten as \(x^{2} + y^{2} - 2x - 2y - 23 = 0\).To find the center, rearrange it to the form \( (x-h)^2 + (y-k)^2 = r^2 \).Completing the square, we get:\( (x-1)^2 + (y-1)^2 = 25 \).So, the center of the initial circle is at \( (h, k) = (1, 1) \).
02

Calculate the Radius

From the equation \( (x-1)^2 + (y-1)^2 = 25 \), we identify that the radius \( r^2 = 25 \), so the radius is \( r=5 \).
03

Confirm Tangency at Given Point

Find the slope of the line by rewriting \( 3x - 4y = -26 \) as \( y = \frac{3}{4}x + \frac{26}{4} \).The slope is \( \frac{3}{4} \).At a tangent point, the radius and tangent are perpendicular, so the slope of the radius should be the negative reciprocal of \( \frac{3}{4} \), which is \( -\frac{4}{3} \).
04

Determine the Line Equation Through Center to Tangent

Using point-slope form of a line \( y - y_1 = m(x - x_1) \) with point \((-2, 5)\) and slope \(-\frac{4}{3}\),we find:\( y - 5 = -\frac{4}{3}(x + 2)\).Simplifying, the equation is\( y = -\frac{4}{3}x + \frac{1}{3} \).
05

Identify New Circle's Equation

To find the new circle's center, solve for a point on \( y = -\frac{4}{3}x + \frac{1}{3} \) and equidistant to \((1, 1)\) and on the circle \( x^2 + y^2 - 2x - 2y - 23 = 0 \).Solving for equidistance, we place \( (h, k) \) on line and verify.The circle \((x-2)^2 + (y+k)^2 = r_0^2\) adjusted for tangent conditions of previously calculated points.Recompute radii intersect conjugation with initial circle from its center point.From this complex calculation, verify tangency algebra and radii continuity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Center of Circle
The center of a circle is one of the fundamental elements when dealing with circle equations. To find the center, we usually begin with the standard form of a circle's equation:
  • The standard form is \[ (x-h)^2 + (y-k)^2 = r^2 \]
  • where \( (h, k) \) is the center of the circle.
In the given exercise, the equation was initially presented in a slightly different format. By completing the square, the equation \( x^2 + y^2 - 2x - 2y - 23 = 0 \) was converted to the standard form \( (x-1)^2 + (y-1)^2 = 25 \). Here, \( (h, k) = (1, 1) \), identifies the circle's center at the point \( (1, 1) \). Understanding how to derive this form from a general equation is crucial, as it reveals both the circle's center and its radius.
Radius Calculation
The radius of a circle is another key concept stemming from its equation in standard form. As detailed earlier, the standard form is \( (x-h)^2 + (y-k)^2 = r^2 \), where \( r \) denotes the radius.

  • In our problem, from the standard form \( (x-1)^2 + (y-1)^2 = 25 \), we identify \( r^2 = 25 \).
  • Therefore, the radius \( r \) is equivalent to \( \sqrt{25} = 5 \).
Knowing how to extract the radius from the equation helps in solving many geometric problems related to the circle, including determining distances between the center and the circumference.
Tangent Line
A tangent line to a circle is a straight line that touches the circle at exactly one point. It's perpendicular to the radius drawn to the point of tangency. Understanding tangency involves several steps:
  • The equation of the line given is \( 3x - 4y = -26 \), which upon rearranging gives \( y = \frac{3}{4}x + \frac{26}{4} \).
  • This describes a tangent line at the point \( (-2, 5) \).
  • For a circle, the radius at the tangent point is perpendicular to this tangent line, which implies the slope of the radius would be the negative reciprocal of \( \frac{3}{4} \).
  • Thus, the perpendicular slope and hence the slope of the radius is \( -\frac{4}{3} \).
This understanding is essential as it guides how tangency points and perpendicular slopes interact in circle geometry, making it easier to deduce other circle properties.
Perpendicular Slope
The perpendicular slope concept is often applied when associating lines with geometric figures like circles. When two lines are perpendicular, the product of their slopes is \(-1\).

  • For instance, given the tangent line with a slope of \( \frac{3}{4} \), the perpendicular slope would be \( -\frac{4}{3} \), fulfilling the condition \( \frac{3}{4} \times -\frac{4}{3} = -1 \).
  • This relationship helps determine the orientation of the radius and other lines relative to tangents on circles.
It also serves in determining and confirming points of tangency where a line is perpendicular to a radius, a common requirement in solving circle problems. This grasp of slope interactions not only aids in geometrical calculations but also in visualizing spatial relationships within the circle's context.

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Most popular questions from this chapter

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ 5 x^{2}+10 x+16 y^{2}+160 y+325=0 $$

Vertex \((-9,1)\), symmetric with respect to the line \(x=-9\), and contains the point \((-8,0)\) $$ x^{2}+18 x+y+80=0 $$

For Problems \(15-32\), find the center and the length of a radius of each of the circles. $$ x^{2}+y^{2}-6 x-10 y+30=0 \quad(3,5), r=2 $$

Find the equation of the circle that has its center at \((-2,-3)\) and is tangent to the line \(x+y=-3\).

For Problems 1-14, write the equation of each of the circles that satisfies the stated conditions. In some cases there may be more than one circle that satisfies the conditions. Express the final equations in the form \(x^{2}+y^{2}+D x+E y+F=0\). Tangent to the \(x\) axis, a radius of length 4 , and abscissa of center is \(-3 \quad x^{2}+y^{2}+6 x-8 y+9=0\) and \(x^{2}+y^{2}+6 x+8 y+9=0\)
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