Problem 101 - A Complete Resource Book in Mathematics for JEE Main

Question

I. The number of common tangents to (A) 3 the circles \(x^{2}+y^{2}-6 x-2 y+9=0\) and \(x^{2}+y^{2}-14 x-8 y+61=0\) is II. The number of common tangents to (B) 4 the circles \(x^{2}+y^{2}=4\) and \(x^{2}+y^{2}-8 x\) \(+12=0\) is III. The number of common tangents to (C) 2 the circles \(x^{2}+y^{2}=4\) and \(x^{2}+y^{2}-6 x\) \(-8 y-24=0\) is IV. The number of tangents to the circle (D) 1 \(x^{2}+y^{2}-8 x-6 y+9=0\) which pass through the point \((3,-2)\) is

Step-by-Step Solution

Answer

A: 0, B: 4, C: 2, D: 1.

Step 1: Rewrite the circle equations in standard form

First, complete the square for the circle equations to express them in the standard form \[(x-h)^2 + (y-k)^2 = r^2\].

Step 2: Circle (A) standard form and analysis

For the first circle \(x^2 + y^2 - 6x - 2y + 9 = 0\), complete the square:- Rearrange: \((x^2 - 6x) + (y^2 - 2y) = -9\).- Complete squares: \((x-3)^2 - 9 + (y-1)^2 - 1 = -9\).- Simplify: \((x-3)^2 + (y-1)^2 = 1\).This represents a circle centered at \((3, 1)\) with radius \(1\).

Step 3: Circle (A) second equation and tangent count

For the second circle \(x^2 + y^2 - 14x - 8y + 61 = 0\), complete the square:- Rearrange: \((x^2 - 14x) + (y^2 - 8y) = -61\).- Complete squares: \((x-7)^2 - 49 + (y-4)^2 - 16 = -61\).- Simplify: \((x-7)^2 + (y-4)^2 = 4\).This circle is centered at \((7, 4)\) with radius \(2\).Calculate distance between centers: \(\sqrt{(7-3)^2 + (4-1)^2} = 5\).The sum of radii is \(3\), so there are no common tangents. Answer is 0.

Step 4: Circle (B) standard forms and tangent calculation

Circle equation \(x^2 + y^2 = 4\) is already in the standard form with center \((0, 0)\) and radius \(2\).For \(x^2 + y^2 - 8x + 12 = 0\), complete the square:- Rearrange: \((x^2 - 8x) + y^2 = -12\).- Complete square: \((x-4)^2 - 16 + y^2 = -12\).- Simplify: \((x-4)^2 + y^2 = 4\).Center is \((4, 0)\), radius \(2\). Distance: \(4\).Identical radii means 4 tangents. Answer is 4.

Step 5: Calculate common tangents to circles in (C)

For circle \(x^2 + y^2 = 4\), the center is at \((0, 0)\), radius \(2\).For \(x^2 + y^2 - 6x - 8y - 24 = 0\),- Rearrange: \((x^2 - 6x) + (y^2 - 8y) = 24\).- Complete square: \((x-3)^2 - 9 + (y-4)^2 - 16 = 24\).- Simplify: \((x-3)^2 + (y-4)^2 = 49\).Center is \((3, 4)\), radius \(7\). Distance: \(5\).Distance between centers < sum of radii (9), so circles intersect: 2 tangents. Answer is 2.

Step 6: Solve part (D) with point-tangent calculation

Circle equation \(x^2 + y^2 - 8x - 6y + 9 = 0\):- Complete the square to convert into \((x-4)^2 + (y-3)^2 = 4\), a circle centered at \((4, 3)\) with radius \(2\).Calculate distance from point \((3, -2)\) to center \((4, 3)\):\(\sqrt{(4-3)^2 + (3+2)^2} = \sqrt{26}\).Only one tangent can pass through, because the point is outside the circle, and only one tangent can be drawn perpendicular to the radius through (3, -2). So, the answer is 1.

Key Concepts

Geometry of Circles

The geometry of circles involves understanding their properties and relationships within a plane. A circle is defined as the set of all points in a plane that are equidistant from a fixed point, known as the center. The distance from the center to any point on the circle is called the radius.

When considering multiple circles, we often explore their interactions by examining common tangents. Common tangents are lines that touch two or more circles at different points.

• If the circles are apart, there can be four tangents (two external and two internal).
• If they touch externally, there are three tangents (two external and one internal).
• If one circle is inside another without touching, there are zero tangents.

This is crucial for understanding how circles relate in a geometric setup, and helps solve problems such as determining the number of common tangents for given circles.

Equation of a Circle

The equation of a circle provides a way to express all the points that make up the circle algebraically. The most common form is the standard form, which is \[ (x-h)^2 + (y-k)^2 = r^2 \]where \( (h, k) \) is the center of the circle and \( r \) is the radius.

This form is particularly useful as it quickly reveals the key properties of the circle:

• The center is easily identified as \( (h, k) \)
• The radius can be read directly as the square root of the constant on the right side.

To convert a general circle equation \\( x^2 + y^2 + Dx + Ey + F = 0 \)to the standard form, complete the square for \\( x \)and \\( y \):
- Rearrange the terms.
- Add and subtract the necessary constants to form perfect squares within the equation, leading to an equation of the standard form.

Radius and Center

The radius and center of a circle are fundamental in describing its size and position. The center of a circle \( (h, k) \) is the point that is equidistant to all points on the circle, while the radius \\( r \) is the fixed distance from the center to the circle's edge.

Determining the radius and center from an equation is straightforward once in the standard form:

• The center can be identified directly as \( (h, k) \)
• The radius is obtained by taking the square root of the constant term on the right-hand side of the standard form equation.

These elements are critical when solving problems involving circles, such as finding intersections, common tangents, or even just understanding the circle's layout on a graph.