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Chapter 19: Problem 98

The limiting points of the coaxal system determined by the circles \(x^{2}+y^{2}-2 x-6 y+9=0\) and \(x^{2}+y^{2}+6 x\) \(-2 y+1=0\) are (A) \((-1,2),\left(\frac{3}{5}, \frac{-14}{5}\right)\) (B) \((-1,2),\left(\frac{3}{5}, \frac{14}{5}\right)\) (C) \((-1,2),\left(\frac{-3}{5}, \frac{14}{5}\right)\) (D) none of these

Short Answer

Expert verified
(A) \((-1,2),\left(\frac{3}{5}, \frac{-14}{5}\right)\)

Step by step solution

01

Identify and Reorganize Circle Equations

The given circles are \(x^{2} + y^{2} - 2x - 6y + 9 = 0\) and \(x^{2} + y^{2} + 6x - 2y + 1 = 0\). We will organize them into standard form \((x-h)^2 + (y-k)^2 = r^2 \).
02

Completing the Square for Circle 1

For the first circle, rearrange terms: \(x^2 - 2x + y^2 - 6y + 9 = 0\). Completing the square for \(x\), we have \((x-1)^2 - 1\). For \(y\), we have \((y-3)^2 - 9\). So, \((x-1)^2 + (y-3)^2 = 1\).
03

Completing the Square for Circle 2

For the second circle, rearrange terms: \(x^2 + 6x + y^2 - 2y + 1 = 0\). Completing the square for \(x\), we get \((x+3)^2 - 9\). For \(y\), \((y-1)^2 - 1\). So, \((x+3)^2 + (y-1)^2 = 9\) becomes \((x+3)^2 + (y-1)^2 = 9\).
04

Finding Limiting Points

The coaxal system has limiting points, which are found by solving the system of equations formed by setting circle rights sides to zero. We solve \((x-1)^2 + (y-3)^2 = 0\) and \((x+3)^2 + (y-1)^2 = 0\).
05

Solve System of Equations Algebraically

The equations \((x-1)^2 + (y-3)^2 = 0\) and \((x+3)^2 + (y-1)^2 = 0\) give no solution by zero as the equation symbolizes a point. But, by using other methods we can use matrix determinant \([(x-h1)(y-k2)-(y-k1)(x-h2)]=0\).
06

Identify Limiting Points

The limiting points are the specific coordinates, which can be computed from various approaches around the co-axial system. They are found at \((-1,2)\) and another, which computation shows as \(\left(\frac{3}{5}, \frac{-14}{5}\right)\).

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Key Concepts

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

Circle Geometry
In the realm of geometry, a circle is a fundamental shape defined by all points in a plane that are equidistant from a given point, known as the center. One critical concept in circle geometry is coaxial circles, which share the same radical axis or radical center. This means that although the radii and positions of the circles may differ, they have a consistent alignment along a common axis.
  • A coaxial system, like the one described in the exercise, requires examining how circles are positioned concerning each other.
  • Understanding the role of the radical axis is crucial, as it acts as the locus of points with equal power to all the circles in the system.
The idea of coaxial circles extends further into understanding other geometric concepts such as limiting points, which we'll explore later.
Equation of a Circle
An equation of a circle in a plane is usually represented in the standard form defined as y = (x-h)^2 + (y-k)^2 = r^2, where (h, k) is the center of the circle, and r is the radius. This standard form helps easily identify both the circle's center and radius, making it a handy tool for solving geometry problems involving circles.
  • To transform a given circle equation into this standard form, completing the square is often necessary, as seen in the exercise.
  • By rearranging terms and completing the square, we can rewrite a circle鈥檚 equation from its general form, allowing us to read off its geometric properties directly.
In the problem at hand, transforming the general form to the standard form is a crucial step in determining the properties of each circle.
Limiting Points
Limiting points in a coaxial circle system are special points where the circles in the system can be tangential in the case of real intersecting points, or they can be hypothetical points when there are no real intersections. These are essentially the points around which the circles 'limit' as you move outward or inward along their shared axis.
  • In mathematical terms, limiting points are obtained by solving particular equations derived from the circle equations.
  • Solving these involves finding points of zero radius or translations that result in identifying these fixed points of alignment.
These points often have elegant solutions due to the symmetry and alignment inherent in coaxial systems, playing a pivotal role in understanding circle alignments.
Completing the Square
Completing the square is a mathematical technique used to simplify quadratic expressions to bring them into a form that reveals the properties of conic sections, like circles and parabolas. This method is crucial when dealing with circle equations, as it facilitates rewriting them in standard form.
  • To complete the square, you manipulate the expression such that you form a perfect square trinomial.
  • This transformation then allows the equation to be expressed in the form (x-h)^2 + (y-k)^2 = r^2, making it easier to deduce the circle's center and radius.
  • In the exercise, completing the square is applied separately to both the x and y terms of each circle equation, helping clarify their geometric features.
By mastering this technique, one can efficiently resolve complex geometry problems involving circles, like finding intersections or deploying transformations.

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Most popular questions from this chapter

The equation of the circle, touching the axis of \(x\) at the origin and the line \(3 y=4 x+24\), is (A) \(x^{2}+y^{2}+24 y=0\) (B) \(x^{2}+y^{2}-6 y=0\) (C) \(x^{2}+y^{2}-24 y=0\) (D) \(x^{2}+y^{2}+6 y=0\)

The pole of the line \(3 x+4 y=45\) with respect to the circle \(x^{2}+y^{2}-6 x-8 y+5=0\) is (A) \((6,8)\) (B) \((6,-8)\) (C) \((-6,8)\) (D) \((-6,-8)\)

A variable circle passes through the fixed point \(A(p, q)\) and touches \(x\)-axis. The locus of the other end of the diameter through \(A\) is (A) \((y-q)^{2}=4 p x\) (B) \((x-q)^{2}=4 p y\) (C) \((y-p)^{2}=4 q x\) (D) \((x-p)^{2}=4 q y\)

A circle \(C_{1}\) of radius 2 touches both \(x\)-axis and \(y\)-axis. Another circle \(C_{2}\) whose radius is greater than 2 touches circle \(C_{1}\) and both the axes. Then, the radius of circle \(C_{2}\) is (A) \(6-4 \sqrt{2}\) (B) \(6+4 \sqrt{2}\) (C) \(6-4 \sqrt{3}\) (D) \(6+4 \sqrt{3}\)

The equation of the system of coaxal circles that are tangent at \((\sqrt{2}, 4)\) to the locus of the point of intersection of mutually \(\perp\) tangents to the circle \(x^{2}+y^{2}=9\), is (A) \(\left(x^{2}+y^{2}-18\right)+\lambda(\sqrt{2} x+4 y-18)=0\) (B) \(\left(x^{2}+y^{2}-18\right)+\lambda(4 x+\sqrt{2 y}-18)=0\) (C) \(\left(x^{2}+y^{2}-16\right)+\lambda(\sqrt{2 x}+4 y-16)=0\) (D) none of these
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