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Chapter 11: Problem 42

If one of the diameters of the circle, given by the equation, \(x^{2}+y^{2}-4 x+6 y-12=0\), is a chord of a circle \(S\), whose centre is at \((-3,2)\), then the radius of \(S\) is: (a) 5 (b) 10 (c) \(5 \sqrt{2}\) (d) \(5 \sqrt{3}\)

Short Answer

Expert verified
The radius of circle S is \(5 \sqrt{2}\).

Step by step solution

01

Identify the Equation of the Circle

The given circle's equation is \(x^{2}+y^{2}-4x+6y-12=0\). To find the center and radius, rewrite it in the standard form \((x-h)^2 + (y-k)^2 = r^2\).
02

Complete the Square

Rewrite the equation by completing the square for both \(x\) and \(y\).\[\begin{align*}x^2 - 4x & : \quad (x-2)^2 = x^2 - 4x + 4,\y^2 + 6y & : \quad (y+3)^2 = y^2 + 6y + 9.\end{align*}\]The equation becomes:\((x-2)^2 + (y+3)^2 = 12+4+9 = 25\).
03

Identify Circle Properties

From the equation \((x-2)^2 + (y+3)^2 = 25\), the center \((h, k)\) is at \((2, -3)\) and the radius \(r\) is \(5\, (\sqrt{25})\).
04

Use Diameter as Chord in Circle S

The circle S has center \((-3, 2)\). Since the diameter equals chord, it passes through \((2, -3)\) and another point symmetrical across the center, say \(P\), making \(d\) symmetrical.
05

Calculate Distance for Symmetrical Point

Find the distance between \((2, -3)\) and the center of \(S\), then mirror to find \(P\). Using distance formula: \[d = \sqrt{(2-(-3))^2 + (-3-2)^2} = \sqrt{25 + 25} = 5\sqrt{2}.\] This is half the chord length, so the diameter is \(2 \times 5\sqrt{2} = 10\sqrt{2}\).
06

Determine the Radius of Circle S

Since the calculated diameter of circle \(S\) is \(10\sqrt{2}\), the radius is half of the diameter, thus \(5\sqrt{2}\).

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Key Concepts

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

Equation of a Circle
In geometry, the equation of a circle is a crucial concept. It defines all the points that make up the shape of the circle on a coordinate plane. A circle is defined by its center \( (h, k) \) and radius \( r \). The general standard form of a circle's equation is given as:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
This equation tells us that any point \( (x, y) \) on the circle is exactly \( r \) units away from the center \( (h, k) \).
  • \( (x - h)^2 \) and \( (y - k)^2 \) represent the squared distances from any point on the circle to the center, along the x and y axes respectively.
  • \( r^2 \) is the square of the radius of the circle.
Understanding this equation allows us to graph the circle and to discern its geometric properties from an algebraic expression.
Completing the Square
Completing the square is a technique used to rewrite a quadratic expression in a form that reveals important geometric properties. It is particularly useful for circles because it helps transform the general form of a circle's equation into the standard form.
Consider a quadratic expression such as \( x^2 - 4x \). The goal is to turn this into a perfect square trinomial. To do this:
  • Take half of the coefficient of \( x \), square it, and add it to \( x^2 - 4x \).
  • For \( x^2 - 4x \), half of \(-4\) is \(-2\), and \( (-2)^2 = 4 \).
  • The expression becomes \( (x - 2)^2 - 4 \).
Repeat this process for \( y \) terms to rewrite each as a square.
This procedure helps simplify and rearrange the equation to readily identify the center and radius of the circle.
Radius Calculation
Calculating the radius of a circle from its equation is straightforward once it's in standard form. Once you have:
\[ (x - 2)^2 + (y + 3)^2 = 25 \]
In this form, \( r^2 = 25 \). Therefore, the radius \( r \) is the square root of \( 25 \), which is \( 5 \).
  • Identify the term on the right side of the equation, which represents the square of the radius.
  • Take the square root of this value to find the radius.
The radius determines the size of the circle and is an essential measurement for understanding the circle's properties.
Diameter as a Chord
In circle geometry, any diameter of a circle can be considered a special type of chord. This becomes significant when a diameter of one circle becomes a chord of another circle. In this scenario, the original circle's diameter runs straight through its center, connecting two points on its edge.
For a circle \( S \) where this diameter acts as a chord, considerations include:
  • Identifying the endpoints of the diameter, which are points on both circles.
  • Understanding that the point of symmetry relative to the center of circle \( S \) means the diameter (or chord) divides the circle into two equal halves.
  • The length of this chord can be calculated using the midpoint formula and the distance formula to confirm its position in space.
Knowing this helps in assessing the length and relevance of the diameters when they function as chords in larger circles.

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Most popular questions from this chapter

The locus of a point which divides the line segment joining the point \((0,-1)\) and a point on the parabola, \(x^{2}=4 y\), internally in the ratio \(1: 2\), is: (a) \(9 x^{2}-12 y=8\) (b) \(9 x^{2}-3 y=2\) (c) \(x^{2}-3 y=2\) (d) \(4 x^{2}-3 y=2\)

If the pair of lines \(a x^{2}+2(a+b) x y+b y^{2}=0\) lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then (a) \(3 a^{2}-10 a b+3 b^{2}=0\) (b) \(3 a^{2}-2 a b+3 b^{2}=0\) (c) \(3 a^{2}+10 a b+3 b^{2}=0\) (d) \(3 a^{2}+2 a b+3 b^{2}=0\)

Let \(\mathrm{C}_{1}\) and \(\mathrm{C}_{2}\) be the centres of the circles \(\mathrm{x}^{2}+y^{2}-2 \mathrm{x}-2 \mathrm{y}-2=0\) and \(x^{2}+y^{2}-6 x-6 y+14=0\) respectively. If \(P\) and \(Q\) are the points of intersection of these circles then, the area (in sq. units) of the quadrilateral \(\mathrm{PC}_{1} \mathrm{QC}_{2}\) is: (a) 8 (b) 6 (c) 9 (d) 4

The straight line \(x+2 y=1\) meets the coordinate axes at \(A\) and \(\mathrm{B}\). A circle is drawn through \(\mathrm{A}, \mathrm{B}\) and the origin. Then the sum of perpendicular distances from \(\mathrm{A}\) and \(\mathrm{B}\) on the tangent to the circle at the origin is : (a) \(\frac{\sqrt{5}}{2}\) (b) \(2 \sqrt{5}\) (c) \(\frac{\sqrt{5}}{4}\) (d) \(4 \sqrt{5}\)

Three distinct points \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are given in the 2-dimensional coordinates plane such that the ratio of the distance of any one of them from the point \((1,0)\) to the distance from the point \((-1,0)\) is equal to \(\frac{1}{3}\). Then the circumcentre of the triangle \(\mathrm{ABC}\) is at the point: (a) \(\left(\frac{5}{4}, 0\right)\) (b) \(\left(\frac{5}{2}, 0\right)\) (c) \(\left(\frac{5}{3}, 0\right)\) (d) \((0,0)\)
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