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Chapter 19: Problem 73

\(S(x, y)=0\) represents a circle. The equation \(S(x, 2)=\) 0 gives two identical solutions \(x=1\) and the equation \(S(1, y)=0\) gives two distinct solutions \(y=0,2\). The equation of the circle is (A) \(x^{2}+y^{2}+2 x+2 y+1=0\) (B) \(x^{2}+y^{2}+2 x+2 y-1=0\) (C) \(x^{2}+y^{2}-2 x-2 y+1=0\) (D) none of these

Short Answer

Expert verified
The circle equation is option (A): \(x^{2} + y^{2} + 2x + 2y + 1 = 0 \).

Step by step solution

01

Identify the General Circle Equation

The general equation for a circle is \( (x-h)^2 + (y-k)^2 = r^2 \) where \((h, k)\) is the center of the circle and \(r\) is the radius. Expanding this gives \( x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0 \).
02

Substitute Values for Line y = 2

Given that \( S(x, 2) = 0 \) gives identical solutions \( x = 1 \), substitute \( y = 2 \) into the general form to get: \[ x^2 + 4 - 2hx - 4k + (h^2 + k^2 - r^2) = 0 \]. Since it has repeated roots at \( x = 1 \), this is a double root condition.
03

Solve the Double Root Condition

For \( x = 1 \) to be a double root, the discriminant must be zero: \[ (-2h)^2 - 4(1)(4 - 4k + (h^2 + k^2 - r^2)) = 0 \] This simplifies to two conditions: 1. Totaling constants gives \( 4-4k + (h^2 + k^2 - r^2)= 4h \), leading to \( k = 1 \).
04

Substitute Values for Line x = 1

Given \( S(1, y) = 0 \) gives distinct solutions \( y = 0, 2 \), substitute \( x = 1 \) into the general form to get: \[ 1 + y^2 - 2h - 2ky + (h^2 + k^2 - r^2) = 0 \] Since the roots are \( y = 0 \) and \( y = 2 \), these are the actual solutions.
05

Solve for Centre Coordinates and Radius

Using conditions from substituting conditions from \( x = 1 \) line, solve: \( y^2 - 4y + (1 + 2 - 2 - 1^2 - h^2 + r^2) = 0 \). Solving with \( y = 0, 2 \) gives the center \( (h, k) = (1, 1) \) and check \( r \) squared gives \( r^2 = 0 \).
06

Complete the Circle Equation

Calculating and simplifying with coordinates, Given equality conditions \( r^2 = 1 \) using both line results,plug in the values: \( h = -1, k = -1, r = \sqrt{1} \) in, so the circle equation is: \( x^2 + y^2 + 2x + 2y + 1\)
07

Verification Against Choices

Matching with provided options: option (A) fits: \( x^2 + y^2 + 2x + 2y + 1=0\). This matches, hence solution is correct.

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Key Concepts

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

Double Root Condition
When dealing with circle equations, encountering a double root condition indicates that the quadratic has a repeated root. In such a case, this occurs when the discriminant of the quadratic expression equals zero. For example, consider the equation given in the problem \( S(x, 2) = 0 \), which has a repeated solution \( x = 1 \). A repeated solution means that when we substitute \( y = 2 \) into the circle's equation, the expression must factor in such a way that one factor is squared. This characteristic indicates a tangent at that specific point of the circle. In this context, a double root signifies that this point lies on the circumference of the circle and is tangent to the line \( y = 2 \). It's a crucial step in identifying specific conditions necessary for finding the exact circle equation.
General Equation for a Circle
The general form of a circle's equation is \( (x-h)^2 + (y-k)^2 = r^2 \), where \((h, k)\) represent the coordinates of the circle's center, and \(r\) is the circle's radius. To express this in standard algebraic form, expand it to \( x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0 \). This form allows us to directly substitute given points or conditions to solve for \(h\), \(k\), and \(r\) to determine the specific circle equation. For instance, in step 2 of the original exercise, by substituting values where \( y = 2 \), we cleverly manipulate this form to discern relationships and solve for the center's coordinates \((h, k)\) and radius \(r\). This method is pivotal in transforming a generalized circle equation to a specific one.
Identifying Circle Center
Determining the center of the circle is crucial when dealing with its equation. By analyzing the transformed equation \( x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0 \), we can identify the circle's center \((h, k)\). From the condition given in the exercise, substituting \( x = 1 \) and concluding that there are two solutions for \( y \) helps us pinpoint the relationship between \(h\) and \(k\). In simpler terms, through solving for these values under the repeated and distinct root conditions, we discover \(h = 1\) and \(k = 1\), accurately placing the circle's center. Furthermore, the process of solving influences not only the coordinates of the center but also determines the correct configuration and size of the circle. Knowing these values guides us to double-check the integration of all components in forming the circle's exact equation.

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Most popular questions from this chapter

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

Assertion: The locus of the centres of circles passing through the origin and cutting the circle \(x^{2}+y^{2}+6 x-\) \(4 y+2=0\) orthogonally is \(3 x-2 y+1=0\). Reason: The two circles \(x^{2}+y^{2}+2 g_{1} x+2 f_{1} y+c_{1}=0\) and \(x^{2}+y^{2}+2 g_{2} x+2 f_{2} y+c_{2}=0\) cut each other orthogonally if \(2 g_{1} g_{2}+2 f f_{2}=c_{1}+c_{2}\)

If the circles \(x^{2}+y^{2}=1\) and \(x^{2}+y^{2}-4 x-6 y+12=0\) cut off equal intercepts on a line which passes through the point \((1,1)\), then the slope of the line is (A) 1 (B) \(-1\) (C) \(\frac{3}{2}\) (D) \(-\frac{3}{2}\)

Two vertices of an equilateral triangle are \((-1,0)\) and \((1,0)\). An equation of its circumcentre is (A) \(x^{2}+y^{2}+\frac{2}{\sqrt{3}} y-1=0\) (B) \(x^{2}+y^{2}-\frac{2}{\sqrt{3}} y-1=0\) (C) \(x^{2}+y^{2}+\frac{2}{\sqrt{3}} y+1=0\) (D) none of these

A tangent drawn from the point \((4,0)\) to the circle \(x^{2}+\) \(y^{2}=8\) touches it at a point \(A\) in the first quadrant. The coordinates of another point \(B\) on the circle such that \(A B=4\), are (A) \((2,-2)\) (B) \((-2,2)\) (C) \((2,2)\) (D) \((-2,-2)\)
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