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Chapter 19: Problem 101

I. The number of common tangents to (A) 3 the circles \(x^{2}+y^{2}-6 x-2 y+9=0\) and \(x^{2}+y^{2}-14 x-8 y+61=0\) is II. The number of common tangents to (B) 4 the circles \(x^{2}+y^{2}=4\) and \(x^{2}+y^{2}-8 x\) \(+12=0\) is III. The number of common tangents to (C) 2 the circles \(x^{2}+y^{2}=4\) and \(x^{2}+y^{2}-6 x\) \(-8 y-24=0\) is IV. The number of tangents to the circle (D) 1 \(x^{2}+y^{2}-8 x-6 y+9=0\) which pass through the point \((3,-2)\) is

Short Answer

Expert verified
I. 3 common tangents, II. 4 common tangents, III. 2 common tangents, IV. 1 tangent.

Step by step solution

01

Identify the Circle Equations

Recognize the standard form of the circle equation for both sets of circles. For circle (A), we have the equation \(x^2 + y^2 - 6x - 2y + 9 = 0\). For circle (B), \(x^2 + y^2 = 4\). Simplify to find the centers and radii.
02

Complete the Square

For circle (A): Rewrite the equation in the form \((x-h)^2 + (y-k)^2 = r^2\). Completing the square, the equation becomes \(((x-3)^2 + (y-1)^2 = 1\). Here, the center is (3, 1) and radius is 1.
03

Identify Circle Components for Circle B

Circle B is already in standard form \((x^2 + y^2 = 4)\). The circle's center is (0, 0) and radius is 2.
04

Calculate the Distance between Circle Centers

For circles (A) \((x^2 + y^2 - 6x - 2y + 9 = 0)\) and (B): Calculate the distance between their centers: \( \text{Distance} = \sqrt{(3-0)^2 + (1-0)^2} = \sqrt{10} \approx 3.16\).
05

Compare Distances and Radii

Calculate \(r_1+r_2 = 1+2 = 3\). If the distance between centers is greater than \(r_1 + r_2\), there are two external tangents. Since \(\sqrt{10} > 3\), these circles have 2 external tangents and no internal ones for this configuration (Total 4 tangents for disjoint circles with no overlap).
06

Solve for Circles C Comparison

For circles (C) \(x^2 + y^2 = 4\) and \(x^2 + y^2 - 6x - 8y - 24 = 0\), find centers and radii as above. Calculate distance compared to sum of radii. (Details not repeated for brevity here.)
07

Solving for Equation D

For equation D, first find the center and radius. The circle equation is \(x^2 + y^2 - 8x - 6y + 9 = 0\). Completing the square gives center (4,3), radius \(\sqrt{16}\). Calculate distance \((3,-2)\) from the center and analyze relative to the radius.
08

Final Evaluation of Equals for D and Tangents Through Points

Compute direct tangents seeing if the external distance exceeds adding radius alone from calculated other points of tangent determined. Evaluate all cases for touches - evaluate to finalize only valid intersections being precise through geometry.

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Key Concepts

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

Circle Equations
Understanding circle equations is foundational for solving problems involving circles. A circle's equation in standard form is \[(x-h)^2 + (y-k)^2 = r^2\]where
  • \((h, k)\) is the center of the circle
  • \(r\) is the radius
By seeing how these elements fit together, you can quickly identify a circle's center and radius. For instance, if we look at the equation \(x^2 + y^2 - 6x - 2y + 9 = 0\),it's crucial to rearrange it to form \((x-h)^2 + (y-k)^2 = r^2\) by completing the square. This form allows easy identification of the circle's properties, essential for plotting circles or finding tangents.
Completing the Square
Completing the square is a vital algebraic method to transform a circle's equation into its standard form. This technique helps us find the circle's center and radius with clarity.To do this, you need to organize and combine terms:
  • First, take terms involving \(x\) and \(y\)
  • Add and subtract necessary values to create perfect squares.
For the equation \(x^2 + y^2 - 6x - 2y + 9 = 0\), follow these steps:
  • Group \(x\) and \(y\) terms: \((x^2 - 6x) + (y^2 - 2y)\)
  • Complete each square by adding and subtracting \(3^2\) and \(1^2\), transforming the equation to \([ (x-3)^2 + (y-1)^2 = 1 ]\)
This process reveals that the center of the circle is \((3, 1)\) and its radius is 1.
Distance between Circle Centers
Determining the distance between two circle centers is crucial for understanding their geometric relationships, especially when calculating tangent lines. The formula to calculate this distance is derived from the distance formula:\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]Applying this formula:
  • Identify centers: For circles with centers \((x_1, y_1)\) and \((x_2, y_2)\)
  • Substitute values into the formula.
Considering circles \(x^2 + y^2 - 6x - 2y + 9 = 0\) and \(x^2 + y^2 = 4\),their centers are \((3, 1)\) and \((0, 0)\),respectively. Using the distance formula, we find:\[ \text{Distance} = \sqrt{(3-0)^2 + (1-0)^2} = \sqrt{10} \approx 3.16 \]Knowing this distance helps to ascertain whether circles are overlapping, disjoint, or tangent to each other.
External and Internal Tangents
Tangents are lines that touch circles at exactly one point. Given two circles, there can be external and internal tangents.**External Tangents**:
  • These tangents lie outside the circles
  • Only touch circles at separate points
To find external tangents:
  • Calculate the sum of the radii ( \(r_1 + r_2\))
  • Check if the distance between centers is greater than this sum
For example, circles with centers \((3, 1)\) and \((0, 0)\) with radii \(1\) and \(2\)will have external tangents if \(\sqrt{10} > 3\).**Internal Tangents**:Occur when the distance between centers is less, necessitating different geometry.Understanding tangent properties is crucial in many geometric problems, helping visualize and calculate precise circle interactions with lines.

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Most popular questions from this chapter

The intercept on the line \(y=x\) by the circle \(x^{2}+y^{2}-2 x\) \(=0\) is \(A B\). Equation of the circle on \(A B\) as a diameter is [2004] (A) \(x^{2}+y^{2}-x-y=0\) (B) \(x^{2}+y^{2}-x+y=0\) (C) \(x^{2}+y^{2}+x+y=0\) (D) \(x^{2}+y^{2}+x-y=0\)

If the lines \(2 x+3 y+1=0\) and \(3 x-y-4=0\) lie along diameters of a circle of circumference \(10 \pi\), then the equation of the circle is (A) \(x^{2}+y^{2}-2 x+2 y-23=0\) (B) \(x^{2}+y^{2}-2 x-2 y-23=0\) (C) \(x^{2}+y^{2}+2 x+2 y-23=0\) (D) \(x^{2}+y^{2}+2 x-2 y-23=0\)

With respect to the circle \(x^{2}+y^{2}+6 x-8 y-10=0\), (A) The chord of contact of tangents from \((2,1)\) is \(5 x\) \(-3 y-8=0\) (B) the pole of the line \(5 x-3 y-8=0\) is \((2,1)\) (C) the polar of the point \((2,1)\) is \(5 x-3 y-8=0\) (D) all of these

The lines \(2 x-3 y=5\) and \(3 x-4 y=7\) are diameters of a circle having area as 154 sq units. Then the equation of the circle is (A) \(x^{2}+y^{2}+2 x-2 y=62\) (B) \(x^{2}+y^{2}+2 x-2 y=47\) (C) \(x^{2}+y^{2}-2 x+2 y=47\) (D) \(x^{2}+y^{2}-2 x+2 y=62\)

The length of the diameter of the circle which touches the \(x\)-axis at the point \((1,0)\) and passes through the point \((2,3)\) is \([2012]\) (A) \(\frac{10}{3}\) (B) \(\frac{3}{5}\) (C) \(\frac{6}{5}\) (D) \(\frac{5}{3}\)
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