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Question

If 3x + y - 5 = 0 is the equation of a chord of the circle x+ y2 - 25 = 0, then what are the coordinates of the mid-point of the chord ?

The correct answer is \( \left(\frac{3}{2}, \frac{1}{2}\right) \)

Finding the Midpoint of a Chord in a Circle

We are given the equation of a circle and the equation of a chord of that circle. Our goal is to find the coordinates of the midpoint of this chord.

Understanding the Circle and the Chord

The equation of the circle is given as \(x^2 + y^2 - 25 = 0\). This can be rewritten as \(x^2 + y^2 = 25\). This is the standard form of a circle centered at the origin \((0,0)\) with a radius squared of 25. So, the center of the circle is \(C = (0,0)\) and the radius is \(r = \sqrt{25} = 5\).

The equation of the chord is given as \(3x + y - 5 = 0\). This is a linear equation representing the line segment (the chord) that intersects the circle.

Geometric Property of the Midpoint of a Chord

A key geometric property relates the center of a circle to the midpoint of any chord: the line segment connecting the center of the circle to the midpoint of the chord is always perpendicular to the chord.

Let the midpoint of the chord be \(M = (h, k)\). The center of the circle is \(C = (0,0)\).

The line segment CM connects \(C(0,0)\) and \(M(h, k)\). The slope of this line segment CM is given by:

\(m_{CM} = \frac{k - 0}{h - 0} = \frac{k}{h}\)

Now, let's find the slope of the chord \(3x + y - 5 = 0\). We can rewrite this equation in the form \(y = mx + c\) to find the slope:

\(y = -3x + 5\)

The slope of the chord is \(m_{chord} = -3\).

Since the line segment CM is perpendicular to the chord, the product of their slopes is -1:

\(m_{CM} \times m_{chord} = -1\)

\(\left(\frac{k}{h}\right) \times (-3) = -1\)

\(\frac{-3k}{h} = -1\)

\(3k = h\)

So, we have a relationship between the coordinates of the midpoint: \(h = 3k\).

Midpoint Lies on the Chord

The midpoint \(M(h, k)\) must also lie on the chord itself. Therefore, the coordinates \((h, k)\) must satisfy the equation of the chord \(3x + y - 5 = 0\). Substituting \((h, k)\) into the chord equation:

\(3h + k - 5 = 0\)

Solving for the Midpoint Coordinates

We now have a system of two linear equations with two variables, \(h\) and \(k\):

  1. \(h = 3k\)
  2. \(3h + k - 5 = 0\)

Substitute the first equation into the second equation:

\(3(3k) + k - 5 = 0\)

\(9k + k - 5 = 0\)

\(10k - 5 = 0\)

\(10k = 5\)

\(k = \frac{5}{10} = \frac{1}{2}\)

Now substitute the value of \(k\) back into the first equation to find \(h\):

\(h = 3k = 3 \times \frac{1}{2} = \frac{3}{2}\)

So, the coordinates of the midpoint of the chord are \(\left(\frac{3}{2}, \frac{1}{2}\right)\).

Summary of Steps

  • Identify the center of the circle from its equation.
  • Find the slope of the given chord.
  • Use the property that the line from the center to the midpoint of the chord is perpendicular to the chord to relate the midpoint's coordinates.
  • Use the property that the midpoint lies on the chord to form another equation.
  • Solve the system of equations to find the midpoint's coordinates.

Revision Table: Circle and Chord Properties

Concept Description Relevance to Problem
Circle Equation (\(x^2 + y^2 = r^2\)) Circle centered at origin \((0,0)\) with radius \(r\). Used to find the center of the given circle.
Chord A line segment connecting two points on the circle. The line \(3x + y - 5 = 0\) is the chord.
Midpoint of Chord The point exactly halfway along the chord. The required output of the problem.
Perpendicular Lines Two lines whose slopes \(m_1, m_2\) satisfy \(m_1 m_2 = -1\). The line from the center to the midpoint is perpendicular to the chord.
Slope of a Line (\(Ax + By + C = 0\)) Given by \(-A/B\). For \(3x + y - 5 = 0\), slope is \(-3/1 = -3\). Used to find the slopes of the chord and the line segment from the center to the midpoint.

Additional Information: Analytical Geometry Concepts

This problem utilizes fundamental concepts from analytical geometry (also known as coordinate geometry).

  • Circle Equations: Understanding standard forms (\((x-a)^2 + (y-b)^2 = r^2\)) helps identify the center \((a,b)\) and radius \(r\). The given circle is a special case centered at the origin.
  • Lines and Slopes: The slope of a line indicates its steepness. The relationship between slopes of perpendicular lines is crucial for solving geometry problems using coordinates.
  • System of Equations: Many coordinate geometry problems lead to a system of equations that need to be solved simultaneously to find unknown coordinates or parameters.
  • Geometric Properties: Translating geometric properties (like perpendicularity) into algebraic equations is key in analytical geometry. The property that the perpendicular from the center of a circle to a chord bisects the chord (meaning it passes through the midpoint) is fundamental here.

Solving problems like this strengthens your understanding of how algebra can be used to describe and solve geometric problems involving circles and lines.

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Important Questions from Circles

  1. What is the area of minor segment ?

  2. What is the area of major segment ?

  3. A straight line x = y + 2 touches the circle 4(x 2+ y 2) = r 2. The value of r is

  4. If the centre of the circle passing through the origin is (3, 4), then the intercepts cut off by the circle on x-axis and y-axis respectively are

  5. What is the radius of the circle passing through the point (2, 4) and having centre at the intersection of the lines x – y = 4 and 2x + 3y + 7 = 0?

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