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Chapter 19: Problem 70

If a chord \(A B\) subtends a right angle at the centre of a given circle, then the locus of the centroid of the triangle \(P A B\) as \(P\) moves on the circle is a/an (A) parabola (B) ellipse (C) hyperbola (D) circle

Short Answer

Expert verified
The locus is a circle.

Step by step solution

01

Understanding the Problem

We have a circle with center O and a chord AB which subtends a right angle (90 degrees) at the center O. We want to find the locus of the centroid of triangle PAB as P moves around the circle.
02

Identifying Key Properties

Since AB subtends a right angle at the center, it's a diameter of the circle. Therefore, the midpoint M of AB is the center O of the circle. That makes triangle PAB a right triangle with the hypotenuse AB when point P moves on the circle.
03

Determining the Centroid Formula

The coordinates of the centroid G of triangle PAB, given vertices at \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), is given by \( (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}) \). In this case, the points are P, A, and B.
04

Setting the Equation for the Circle

Assume the circle has radius r and its center O is at the origin (0,0). Points A and B lie on the circle and form the endpoints of the diameter. Let A and B have coordinates \((r, 0)\) and \((-r, 0)\), and P has coordinates (x, y).
05

Substitute and Simplify

The centroid G of PAB is given by \(G = (\frac{x + r + (-r)}{3}, \frac{y + 0 + 0}{3}) = \left( \frac{x}{3}, \frac{y}{3} \right)\). Since point P is on the circle, \(x^2 + y^2 = r^2\).
06

Equation for Locus of the Centroid

If \(G = \left( \frac{x}{3}, \frac{y}{3} \right)\), then \((3X)^2 + (3Y)^2 = r^2\), where \(G = (X, Y)\) is the centroid. This simplifies to \(9X^2 + 9Y^2 = r^2\) or \(X^2 + Y^2 = \frac{r^2}{9}\).
07

Identify the Locus

The equation \(X^2 + Y^2 = \frac{r^2}{9}\) represents a circle with radius \(\frac{r}{3}\). Thus, the locus of the centroid as P moves around the circle is also a circle.

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Key Concepts

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

Chord Subtends Right Angle
When a chord of a circle subtends a right angle at the center of the circle, it indicates a special relationship. If the chord is AB and it subtends a 90-degree angle at the circle's center, this implies that the chord is actually a diameter. This is because only a diameter can span across and create such an angle with the center, thanks to the properties of a circle.
  • A diameter is the longest chord in a circle.
  • Every diameter subtends a right angle (90 degrees) at the circle's center.
  • This property is essential to determine loci and geometric relationships.

The diameter property ensures that no matter which point P lies on the circle, triangle PAB will always be a right triangle. This understanding is crucial as it defines how the loci of related geometric figures are established.
Circle Geometry
Circle geometry is a fundamental part of understanding many geometric concepts and problems. It involves the study and application of various elements like chords, diameters, radii, and angles within the context of a circle. In our exercise, we observe several important concepts of circle geometry:
  • Circle Center: The fixed point which defines the set of all points on the circle equidistant from this center.
  • Radius: The distance from the center to any point on the circle.
  • Diameter: A special chord that passes through the center and is twice the length of the radius.
  • Chord: A line segment whose endpoints lie on the circle, and when it subtends a right angle, it signifies specific loci relationships.

Understanding the placement of points and lines within a circle is key to figuring out the locus and the shape of paths traced by moving points, such as the centroid of a triangle in circle-related exercises like these.
Triangle Centroid Formula
The centroid of a triangle, sometimes called the triangle's center of gravity, is the point where its three medians intersect. Each median is a line segment connecting a vertex to the midpoint of the opposite side. This holds true in triangle PAB as point P moves on the circle.
The formula to find the centroid of a triangle with vertices at points oids (\(x_1, y_1\)), \((x_2, y_2\)), \((x_3, y_3\)) is:
\[G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\]
  • Centroid computation: Crucial for determining the locus of the centroid as a point moves on the circle.
  • The locus equation: For our exercise, transforming the equation based on changing positions of P helps find that the locus itself is a smaller circle, with the locus circle's radius being one-third of the original circle.
This geometric approach using centroid calculations can simplify a lot of common geometric problems, providing insights into the movement and positions of points and lines.

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Most popular questions from this chapter

If a circle passes through the points where the lines \(3 k x-2 y-1=0\) and \(4 x-3 y+2=0\) meet the coordinate axes then \(k=\) (A) 1 (B) \(-1\) (C) \(\frac{1}{2}\) (D) \(\frac{-1}{2}\)

The locus of the centre of a circle which passes through the point \((0,0)\) and cuts off a length \(2 b\) from the line \(x\) \(=c\), is (A) \(y^{2}+2 c x=b^{2}+c^{2}\) (B) \(x^{2}+c x=b^{2}+c^{2}\) (C) \(y^{2}+2 c y=b^{2}+c^{2}\) (D) none of these

The circle \(x^{2}+y^{2}=4 x+8 y+5\) intersects the line \(3 x\) \(-4 y=m\) at two distinct points then \(\mathrm{m}\) satisfies [2010] (A) \(-35

The circle \(x^{2}+y^{2}-4 x-4 y+4=0\) is inscribed in a triangle which have two of its sides along the coordinate axes. If the locus of the circumcentre of the triangle is \(x+y-x y+k \sqrt{x^{2}+y^{2}}=0\), then \(k\) is equal to (A) 1 (B) \(-1\) (C) 2 (D) none of these

If the circle \(x^{2}+y^{2}+2 g x+2 f y+c=0\) bisects the circumference of the circles \(x^{2}+y^{2}+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0\), then (A) \(2 g^{\prime}\left(g-g^{\prime}\right)+2 f^{\prime}\left(f-f^{\prime}\right)=c-c^{\prime}\) (B) \(g^{\prime}\left(g-g^{\prime}\right)+f^{\prime}\left(f-f^{\prime}\right)+c-c^{\prime}=0\) (C) \(2 g\left(g-g^{\prime}\right)+2 f\left(f-f^{\prime}\right)=c-c^{\prime}\) (D) none of these
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