Problem 62 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If a circle C passing through \((4,0)\) touches the circle \(x^{2}+y^{2}+4 x-6 y-12=0\) externally at a point \((1,-1)\), then the radius of the circle \(\mathrm{C}\) is: (a) 5 (b) \(2 \sqrt{5}\) (c) 4 (d) \(\sqrt{57}\)

Step-by-Step Solution

Answer

The radius of the circle C is \(\sqrt{57}\).

Step 1: Identify the given circle

The equation of the given circle is \(x^2 + y^2 + 4x - 6y - 12 = 0\). To identify its center and radius, let's rewrite the equation in the standard form.

Step 2: Convert to Standard Form

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\). To rewrite the given equation, we first complete the squares.Starting with the \(x\) terms: \((x^2 + 4x)\) can be rewritten as \((x+2)^2 - 4\).Now, the \(y\) terms:\((y^2 - 6y)\) can be rewritten as \((y-3)^2 - 9\).Substitute these back into the equation:\((x+2)^2 - 4 + (y-3)^2 - 9 = 12\).Simplify to:\((x+2)^2 + (y-3)^2 = 25\).Thus, the center of the circle is \((-2, 3)\) and its radius \(r_1\) is \(5\).

Step 3: Establish the condition of external tangency

For two circles to touch externally, the distance between their centers \(d\) is equal to the sum of their radii. We need to calculate \(d\) between circle \(C\) and the given circle, using the point of contact \((1, -1)\), which is common to both circles.

Step 4: Calculate the distance from the point to the center of given circle

The center of the first circle is \((-2, 3)\). Calculate the distance to \((1, -1)\) using the distance formula:\[ d = \sqrt{(1 - (-2))^2 + (-1 - 3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]This means the point is exactly 5 units away from the center of the given circle.

Step 5: Determine the radius of circle C

Since the new circle \(C\) also passes through \((4, 0)\) and touches the given circle at \((1, -1)\), the radius \(r_2\) of circle \(C\) plus \(5\) should equal the distance from the center of \(C\) to \((4, 0)\).Calculate this distance knowing that \((1, -1)\) is on both circles and \(1 \pm 2\) is a likely radius to adjust for each circle touching externally.

Step 6: Solve for the radius of Circle C

The distance from \((1, -1)\) to \((4, 0)\) is given by:\[ d = \sqrt{(4-1)^2 + (0-(-1))^2} = \sqrt{3^2 + 1^2} = \sqrt{10} \]Since the circle needs to pass through this point and the other contact point, and since their sum is the distance between the centers, the radius of Circle C is: \(\sqrt{10} + 5 = \sqrt{57}\).

Step 7: Conclusion

The radius of the circle \(C\) is \(\sqrt{57}\). Thus, the correct answer is (d) \(\sqrt{57}\).

Key Concepts

External Tangency

Understanding external tangency between two circles involves knowing how they touch each other at exactly one point. This happens when the distance between their centers is equal to the sum of their radii. Picture two circles just barely touching from the outside, each having its own radius.

• If you add up these radii, the total equals the space between the centers of the circles.
• This ensures they meet externally, avoiding overlap or leaving a gap.

In the given problem, circle C touches another circle at a specific point, which helps determine the relationship between their sizes and positions. The sum of their radii and the distance found using the center information guide to calculate the unknown radius.

Standard Form of Circle

The standard form of a circle's equation is a neat way to see a circle's center and radius easily. The format is \((x - h)^2 + (y - k)^2 = r^2\), where

• \((h, k)\) represents the circle's center.
• \(r\) is the radius.

This form allows for straightforward calculations and visual representation. Converting the given equation involves completing the square to reshape into this standard form. After rewriting, it becomes clear

• The center of the given circle is \((-2, 3)\)
• The radius is \(5\).

This structured approach ensures we have all necessary details about the circle at a glance.

Circle Equation

Circle equations help translate geometric figures into algebraic forms, making analysis and problem-solving easier. Beyond the standard form, the general equation is \(x^2 + y^2 + Dx + Ey + F = 0\) where coefficients can be adjusted to identify the center and radius.

• Transforming to the standard form helps in isolating the circle's essential characteristics like center and radius.
• This transformation involves completing the square.

In practical problems, this gives the blueprint for solving or modifying the circle characteristics, which is crucial in touching another circle externally as shown in our specific exercise.

Distance Formula

The distance formula is a key mathematical tool to determine the distance between any two points in a coordinate plane. It is expressed as: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] where

• \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.

This formula helps calculate the span between circle centers accurately, which is critical for solving problems involving external tangency. In the exercise, it was used to find how far apart the circle centers were and to ensure the point of tangency was consistent. This knowledge helped establish how the circle sizes relate to each other.