Problem 51 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A triangle has two of its sides along the axes. If the third side touches the circle \(x^{2}+y^{2}-2 a x-2 a y+a^{2}=\) 0 , then the equation of the locus of the circumcentre of the triangle is (A) \(2 a(x+y)=2 x y+a^{2}\) (B) \(2 a(x-y)=2 x y+a^{2}\) (C) \(2 a(x+y)=2 x y-a^{2}\) (D) none of these

Step-by-Step Solution

Answer

(A) \(2a(x+y)=2xy+a^2\) is the correct answer.

Step 1: Understand the Problem

You need to determine the equation of the locus of the circumcenter of a triangle given that two sides are along the coordinate axes and the third side is tangent to the circle \( x^2 + y^2 - 2ax - 2ay + a^2 = 0 \).

Step 2: Analyze the Circle

Notice that the equation \( x^2 + y^2 - 2ax - 2ay + a^2 = 0 \) can be rewritten in a completed square form to find the circle's center and radius. Completing the square, we obtain: \((x - a)^2 + (y - a)^2 = 0\). This shows that the center of the circle is \((a, a)\) and the radius is 0, meaning it is just a point at \((a, a)\).

Step 3: Determine the Triangle's Characteristics

Since the triangle's sides are along the axes, its right-angle vertex is at the origin. Let the point on the axes be \((b, 0)\) on the x-axis and \((0, c)\) on the y-axis, with the hypotenuse touching the point \((a, a)\). So the hypotenuse equates \( y = -\frac{c}{b} x + c\).

Step 4: Find the Tangency Point

The tangency condition is that the point \((a, a)\) satisfies the equation of the hypotenuse. Substitute \(a\) into the equation: \(a = -\frac{c}{b}a + c\). Simplify to get \(b = c\), implying that the slopes are equal.

Step 5: Derive the Locus of the Circumcenter

The circumcenter of the triangle formed by \((b,0)\), \((0,b)\), and \((0,0)\) is at \( (\frac{b}{2}, \frac{b}{2})\) because it is the midpoint of the hypotenuse. Since \(b\) is the quantity that changes as the triangle changes while being tangent to the point \((a, a)\), let \(b = a\). Therefore, the circumcenter takes the form \((\frac{a}{2}, \frac{a}{2})\). The locus of such points \((x, y)\) is \(2ax = 2xy + a^2\).

Step 6: Verify the Options

From our derivation, the correct equation is \(2a(x+y)=2xy+a^2\). Compare with the given options: Option (A) \(2 a(x+y)=2 x y+a^{2}\) is the correct equation.

Key Concepts

Tangent to Circle

A line is tangent to a circle when it touches the circle at exactly one point. This contact point is called the point of tangency. For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must be equal to the circle's radius.
In the given exercise, the circle equation is converted to completed square form, showing that it represents a single point at \((a, a)\). Therefore, the line representing the triangle's hypotenuse must pass through exactly this point with zero radius, meeting the tangency condition.

Completed Square Form

To understand how a quadratic equation represents a circle, it’s often helpful to complete the square, which transforms the equation into a clearer format. The standard circle equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) the radius.
Take the equation \(x^2 + y^2 - 2ax - 2ay + a^2 = 0\). By completing the square, we rewrite it as \((x - a)^2 + (y - a)^2 = 0\), revealing that the figure is not a traditional circle nor a circle of fact but a single point at \((a, a)\). The radius is zero, meaning the circle becomes a point.

Equation of Hypotenuse

For a triangle with legs along the axes, the hypotenuse is simply the line connecting these legs. Its equation can be formulated based on simple geometry. Given points \((b, 0)\) on the x-axis and \((0, c)\) on the y-axis, the hypotenuse is described by the equation \(y = -\frac{c}{b}x + c\).
Here, the slope \(-\frac{c}{b}\) is determined by the relative positions of \(b\) and \(c\), while \(c\) is the y-intercept, indicating where the line cuts the y-axis.
To fulfill tangency, the line, or hypotenuse, goes through point \((a, a)\), hence forcing the condition \(b = c\), simplifying the consideration of the problem.

Triangles Along Axes

Triangles with sides on the x and y axes create right triangles. The right-angle vertex lies at the origin \((0, 0)\). The interest lies in the third side or hypotenuse, influenced by points along these axes.
Imagine a triangle with vertices \((0, 0)\), \((b, 0)\), and \((0, b)\). The hypotenuse intersects at the line \(y = -\frac{c}{b}x + c\). When the hypotenuse touches a circle tangentially, such as point \((a, a)\) in this exercise, it determines the triangle's form and influences the locus of its circumcenter.
Calculating this involves balancing these conditions to find where the circumcenter of such a triangle might shift as \((b, c)\) change while maintaining tangency, leading to the derived locus equation.