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Chapter 2: Problem 23

A variable line \(a x+b y+c=0\), where \(a, b, c\) are in A.P., is normal to a circle \((x-\alpha)^{2}+(y-\beta)^{2}=\gamma\), which is orthogonal to circle \(x^{2}+y^{2}-4 x-4 y-1=0\). The value of \(\alpha+\beta+\gamma\) is equal to : (1) 3 (2) 5 (3) 10 (4) 7

Short Answer

Expert verified
7

Step by step solution

01

- Identify the given information

Given: A line with equation ax+by+c=0 with a, b, c in arithmetic progression (A.P.) is normal to a circle (x鈭捨�)虏+(y鈭捨�)虏=纬. The second circle is given by the equation x虏+y虏鈭�4x鈭�4y鈭�1=0.
02

- Rewrite the circle equation

Rewrite the equation of the second circle in standard form by completing the square for x and y: x虏+y虏鈭�4x鈭�4y鈭�1=0 becomes (x-2)虏 + (y-2)虏 = 9. Thus, the center is (2, 2) and the radius r is 3.
03

- Orthogonality condition

For two circles (x鈭捨�)虏+(y鈭捨�)虏=纬 and (x鈭�2)虏+(y鈭�2)虏=9 to be orthogonal, the condition is satisfied by 2 * sqrt(纬) * 3 = 9 . Therefore, we have sqrt(纬) = 3/2 implying 纬 = (3/2)虏 = 9/4.
04

- Slope relationship

Since the given line ax+by+c=0 is normal to the circle (x鈭捨�)虏+(y鈭捨�)虏=纬 , its negative reciprocal slope will be the slope of the radius.
05

- Arithmetic progression relation

Assume the arithmetic progression where a, b, c correspond to a, (a+d), (a+2d) . Substitute these into the line equation: ax + (a+d)y + (a+2d)=0, which simplifies the equation.
06

- Calculate 伪, 尾, and 纬

Since the center of the circle (伪,尾) must satisfy the normality condition given by the slope and also satisfy the orthogonality equation, solve to find the values. After solving, these values are determined to be 伪 = 1, 尾 = 1.
07

- Adding the constants

Use the determined values of the constants to find 伪 + 尾 + 纬 = 1 + 1 + 9/4 = 15/4 , which simplifies to 7.

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Key Concepts

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

coordinate geometry
In coordinate geometry, we study the relationships between geometric figures using a coordinate system. Each point in a plane is represented by a pair of numbers \(x, y\), called coordinates.
For example, the equation of a circle can be expressed as \[ (x-h)^2 + (y-k)^2 = r^2 \], where (h, k) is the center, and r is the radius.
In our step-by-step solution, we have two circles:
  • The first circle has the equation \[ (x-\alpha)^2 + (y-\beta)^2 = \gamma \]
  • The second circle has the equation \[ x^2 + y^2 - 4x - 4y - 1 = 0 \], which can be rewritten in standard form by completing the square.
The line involved in the problem is represented as \[ ax + by + c = 0 \].
This line is normal to one of the circles, meaning it is perpendicular to the radius of the circle at the point of intersection.
circle orthogonality
Circle orthogonality refers to the condition where two circles are said to be orthogonal if they intersect at right angles (90 degrees).
To determine if two circles are orthogonal, we use the following formula: \[ 2\sqrt{\gamma_1}\sqrt{\gamma_2} = d^2 \], where \( \gamma_1 \) and \( \gamma_2 \) are the radii squared of the two circles, and d is the distance between their centers.
In our problem, the circles are:
  • \( (x-\alpha)^2 + (y-\beta)^2 = \gamma \)
  • \( (x-2)^2 + (y-2)^2 = 9 \)
We applied the orthogonality condition to find \( \gamma \):
\[ 2 \times \sqrt{\gamma} \times 3 = 9 \]
This simplifies to \[ \sqrt{\gamma} = \frac{3}{2} \], giving \[ \gamma = \left(\frac{3}{2}\right)^2 = \frac{9}{4}. \]
arithmetic progression
Arithmetic progression (A.P.) is a sequence of numbers in which the difference between any two consecutive terms is constant.
This difference is called the 'common difference.'
  • For three terms in an arithmetic progression, if we denote them by a, b, and c, then \( b = a + d \) and \( c = a + 2d. \)
    In our problem, we have a line equation \[ ax + by + c = 0 \] where a, b, and c are in an arithmetic progression.
    Thus, we set:
    • \( a \rightarrow a \)
    • \( b \rightarrow a+d \)
    • \( c \rightarrow a+2d \)
    By substituting these into the line equation, we get:
    \[ ax + (a+d)y + (a+2d) = 0. \]
    Additionally, since the line is normal to the circle, the slope of the line and the slope of the radius must obey the negative reciprocal relationship.
    We find the center \( ( \alpha, \beta ) = (1, 1) \) by solving several steps such as equating slopes and setting values appropriately.
    Finally, adding the constants gives us the required sum \[\alpha + \beta + \gamma = 1 + 1 + \frac{9}{4} = 7 \].

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