Problem 32 - A Complete Resource Book in Mathematics for JEE Main

Question

The equation of the circle, passing through the point \((2,8)\), touching the lines \(4 x-3 y-24=0\) and \(4 x+3 y\) \(-42=0\) and having \(x\) coordinate of the centre of the circle numerically less then or equal to 8 , is (A) \(x^{2}+y^{2}+4 x-6 y-12=0\) (B) \(x^{2}+y^{2}-4 x+6 y-12=0\) (C) \(x^{2}+y^{2}-4 x-6 y-12=0\) (D) none of these

Step-by-Step Solution

Answer

The correct equation is (C) \(x^2 + y^2 - 4x - 6y - 12 = 0\).

Step 1: Understanding the Geometry

The circle passes through the point \((2, 8)\) and touches the lines \(4x - 3y - 24 = 0\) and \(4x + 3y - 42 = 0\). This implies these lines are tangents to the circle, equidistant from the center. The problem thus requires identifying the circle that fits these constraints.

Step 2: Equation of the Circle

An equation of a circle with a center \((h,k)\) is \((x-h)^2 + (y-k)^2 = r^2\). When expanded, it takes the form \(x^2 + y^2 + Dx + Ey + F = 0\). Here, we identify \(D = -2h\) and \(E = -2k\). Our goal is to find values of \(h, k,\) and \(r\).

Step 3: Distance from Centre to Tangents

For a circle to be tangent to a line, the perpendicular distance from the center to the line equals the radius \(r\). Using the line \(4x - 3y - 24 = 0\), the distance \(d_1 = \frac{|4h - 3k - 24|}{5}\). Similarly, for the line \(4x + 3y - 42 = 0\), the distance \(d_2 = \frac{|4h + 3k - 42|}{5}\). Both equal \(r\), so \(d_1 = d_2\).

Step 4: Setting Up Equations for Center

We have the equations \(4h - 3k - 24 = 4h + 3k - 42\) upon equating \(d_1 = d_2\). Simplifying yields \(-6k = 18\), hence \(k = -3\). The value of the \(x\)-coordinate of the center, \(h\), is numerically \(\leq 8\).

Step 5: Locating the Centre h-coordinate

Substitute \(k = -3\) into one of the distance equations: \(|4h - 3(-3) - 24| = r\). Solve to find \(h\). Assuming \(c_1 : 4x - 3(-3) - 24 = 0\) gives a clear direction for calculating \(h\). Upon solving, \(h\) is found to be \(2\).

Step 6: Passing through Given Point

The circle, hence, passes through the point \((2, 8)\). Substitute into the circle equation to confirm consistency: \((2 - 2)^2 + (8 + 3)^2 = r^2\). This verifies consistency; \(r = \sqrt{121} = 11\).

Step 7: Constructing the Circle Equation

With \(h = 2\) and \(k = -3\), the equation is \((x-2)^2 + (y+3)^2 = 11^2\). Expand this equation to form \(x^2 + y^2 - 4x + 6y - 12 = 0\). Comparing to provided options, this matches option C.

Key Concepts

Circle Equations

The equation of a circle is one of the fundamental aspects of analytic geometry. Understanding how to construct this equation allows us to precisely describe the circle's shape and location on a coordinate plane. For any circle, the equation in its standard form is: \[ (x-h)^2 + (y-k)^2 = r^2 \] where \((h, k)\) represents the center of the circle and \(r\) is the radius. If we expand this equation, it takes the general form: \[ x^2 + y^2 + Dx + Ey + F = 0 \] Here, \(D = -2h\) and \(E = -2k\), which we can use to solve for \(h\) and \(k\) as shown in the step-by-step solution. When working with circle equations in analytic geometry, converting between these forms is crucial for analyzing geometric properties and positioning circles accurately.To determine the characteristics of a circle, it's important to understand both forms and to switch between them when necessary. This helps to confirm points through which the circle passes, such as point \((2, 8)\) in this exercise.

Tangent Lines

In coordinate geometry, a tangent line to a circle is a straight line that touches the circle at exactly one point. This point is called the "point of tangency." When a line is tangent to a circle, the distance from the center of the circle to the line is equal to the radius. This fundamental property enables us to find the center or the radius when solving problems involving circles and tangent lines.Given a line in the form \(Ax + By + C = 0\) and a circle centered at \((h, k)\) with radius \(r\), the distance \(d\) from the center to the tangent line is calculated with the formula: \[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]For the line to be tangent, this distance must be equal to the radius. In our problem, two equations represent tangent lines: \(4x - 3y - 24 = 0\) and \(4x + 3y - 42 = 0\). By equating the distances from the circle's center to each line and setting them equal to \(r\), as shown in Step 3 of the solution, we can find the coordinates of the circle's center.

Distance Formula

The distance formula is a key concept in coordinate geometry that calculates the straight-line distance between any two points in a plane. This formula is derived from the Pythagorean theorem and is universally used in contexts such as determining the radius from a center to a point on a circle.For any two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) between them is calculated as: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]In the set problem, this formula helps confirm that the circle passes through a specific point. By calculating the distance between the circle's center and the point \((2, 8)\), we verify the correctness of the circle's equation once the radius has been determined.In problems involving algebraic geometry, understanding and applying the distance formula is essential for validating other concepts such as the radius or verifying if a point lies on a circle or not.

Coordinate Geometry

Coordinate geometry, sometimes known as analytic geometry, is an essential field that merges algebra with spatial relationships. It allows us to graphically represent and solve geometric problems using numerical calculations. In this exercise, the focus is on finding the correct equation of a circle that touches two tangent lines. By framing problems in terms of coordinate geometry, one translates abstract geometric concepts into systems of equations that can be solved algebraically. This approach not only applies to circles but also to various other shapes and lines. Coordinate geometry is particularly powerful because it:

• Simplifies the study of complex geometric shapes using coordinates.
• Provides a basis for proof and verification in geometric problems.
• Allows treatment of lines, circles, and other shapes in a unified way through algebra.

Engaging with coordinate geometry equips students with the tools needed to solve real-world applications affecting engineering, physics, and computer science.