Problem 82 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

For the circles \(S_{1} \equiv x^{2}+y^{2}-4 x-6 y-12=0\) and \(S_{2} \equiv\) \(x^{2}+y^{2}+6 x+4 y-12=0\) and the line \(L \equiv x+y=0\) (A) \(L\) is the common tangent of \(S_{1}\) and \(S_{2}\) (B) \(L\) is the common chord of \(S_{1}\) and \(S_{2}\) (C) \(L\) is radical axis of \(S_{1}\) and \(S_{2}\) (D) \(L\) is perpendicular to the line joining the centres of \(S_{1}\) and \(S_{2}\)

Step-by-Step Solution

Answer

(C) L is the radical axis of S1 and S2.

Step 1: Solve for Centers and Radii

For circle \( S_1: x^2 + y^2 - 4x - 6y - 12 = 0 \), complete the square to find the center and radius. Rewrite it as \((x-2)^2 + (y-3)^2 = 25\). Thus, the center is \((2, 3)\) and the radius is \(5\).For circle \( S_2: x^2 + y^2 + 6x + 4y - 12 = 0 \), rewrite it as \((x+3)^2 + (y+2)^2 = 25\). Thus, the center is \((-3, -2)\) and the radius is \(5\).

Step 2: Find Distance Between Centers

Calculate the distance between the centers \((2, 3)\) and \((-3, -2)\) using the distance formula.The distance is calculated as \[\sqrt{(2 - (-3))^2 + (3 - (-2))^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}\]

Step 3: Analyze Line L

The line \( L : x + y = 0 \) is a line with slope \(-1\) going through the origin \((0,0)\).We now assess which role this line plays in relation to the circles \( S_1 \) and \( S_2 \).

Step 4: Check if L is the Common Tangent

For \( L \) to be a common tangent to both circles, the perpendicular distance from the center of each circle to the line \( L \) should equal their radius.Distance from \((2, 3)\) to \( L \): \[ \frac{|2 + 3|}{\sqrt{1^2 + 1^2}} = \frac{5}{\sqrt{2}} \]Distance from \((-3, -2)\) to \( L \): \[ \frac{|-3 - 2|}{\sqrt{1^2 + 1^2}} = \frac{5}{\sqrt{2}} \]Both distances are approximately \(3.54\), which should equal the radius \(5\) for both circles; this is not the case, therefore \( L \) is not a common tangent.

Step 5: Check if L is a Radical Axis

The radical axis is given by the line obtained from subtracting the equations of \( S_1 \) and \( S_2 \).Subtracting: \((x^2 + y^2 - 4x - 6y - 12) - (x^2 + y^2 + 6x + 4y - 12) = 0\)\[-10x - 10y = 0\]This simplifies to \( x + y = 0 \), which is exactly the line \( L \). Hence, \( L \) is indeed the radical axis.

Step 6: Conclusion

The analysis shows that \( L \) is the radical axis of \( S_1 \) and \( S_2 \), as verified by subtracting their equations which results in \( L \). Also check for common chord and perpendicularity to centers failed as well based on computation above.

Key Concepts

circle equations

Circle equations are essential when defining the geometric properties of a circle on a Cartesian plane. A standard equation for a circle is given by \[(x-h)^2 + (y-k)^2 = r^2\]where

• \(h\) and \(k\) are the coordinates of the center of the circle, and
• \(r\) is the radius.

To find the center and radius of a circle equation like \(x^2 + y^2 - 4x - 6y - 12 = 0\), we rearrange into this standard form by completing the square:

• Group the \(x\) terms and \(y\) terms separately.
• Add and subtract constants to make them perfect squares.

For example, by rearranging \((x-2)^2 + (y-3)^2 = 25\), the center is \((2,3)\), and the radius is \(5\). This process helps to visualize and understand circle geometries easily.

distance formula

To find the distance between two points, say \((x_1, y_1)\) and \((x_2, y_2)\), on the coordinate plane, we use the distance formula:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This formula originates from the Pythagorean theorem, making it an invaluable tool to calculate distances between any two points efficiently. In our exercise, finding the distance between the centers of two circles \((2,3)\) and \((-3,-2)\) is key to several analyses.

• Plugging these coordinates into the formula yields \(\sqrt{50} = 5\sqrt{2}\).
• This distance helps assess whether certain lines serve as tangents or chords.

Remember, regardless of how complex a geometry problem, start by calculating key distances to build a deeper understanding.

common tangent

Focusing on the concept of a common tangent involves understanding lines that touch circles at exactly one point. For a line to be a common tangent to two circles, the perpendicular distance from the center of each circle to the line must equal the circle's radius. For two circles, like in our scenario, take circle centers \((2,3)\) and \((-3,-2)\) with radii \(5\). A line \(x+y=0\) is considered a tangent only if:

• The perpendicular distances from both centers are exactly their radii.

Upon calculating using \[\text{distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}\],one finds the distances do not match the radii. Thus, \(x+y=0\) is not a tangent. Understanding tangents is crucial for solving problems involving more than one circle and is foundational in many geometry applications.

perpendicular lines

Perpendicular lines intersect at a right angle of \(90^\circ\). In a coordinate plane, two lines are perpendicular if the product of their slopes is \(-1\). With line \(x+y=0\), you see it’s slope is \(-1\). Now consider the line joining centers of the circles, \(y_2-y_1\) over \(x_2-x_1\):

• The slope between centers \((2,3)\) and \((-3,-2)\) is \([(-2)-3] / [(-3)-2]=1\).

Clearly, since \(-1 \times 1 = -1\), the line \(x+y=0\) is perpendicular. This insight simplifies resolving geometry problems by offering a direct route when determining relationships between lines and shapes.