Problem 27 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the line \(3 x+a y-20=0\) cuts the circle \(x^{2}+y^{2}=25\) at real, distinct or coincident points, then \(a\) belongs to the interval (A) \([-\sqrt{7}, \sqrt{7}]\) (B) \((-\sqrt{7}, \sqrt{7})\) (C) \((-\infty-\sqrt{7}] \cup[\sqrt{7}, \infty)\) (D) none of these

Step-by-Step Solution

Answer

Option (C): \((-\infty, -\sqrt{7}] \cup [\sqrt{7}, \infty)\).

Step 1: Identify the Intersection Condition

For a line to intersect a circle, the perpendicular distance from the center of the circle to the line must be less than or equal to the radius of the circle. The center of the given circle \(x^2 + y^2 = 25\) is \((0, 0)\) and the radius is 5.

Step 2: Perpendicular Distance Formula

The formula for the perpendicular distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\). Substitute \((x_1, y_1) = (0, 0)\) and the line \(3x+ay-20=0\).

Step 3: Apply the Distance Condition

Here the perpendicular distance \(d\) must satisfy \(d \leq 5\). Calculate the distance: \[ d = \frac{|0 + 0 - 20|}{\sqrt{3^2 + a^2}} = \frac{20}{\sqrt{9 + a^2}}. \] So, set \(\frac{20}{\sqrt{9 + a^2}} \leq 5\).

Step 4: Solve the Inequality

Simplify and solve the inequality: \( \frac{20}{\sqrt{9 + a^2}} \leq 5 \Rightarrow \frac{20}{5} \leq \sqrt{9 + a^2} \Rightarrow 4 \leq \sqrt{9 + a^2}\).

Step 5: Square Both Sides

Square both sides of the inequality to eliminate the square root: \( 16 \leq 9 + a^2 \Rightarrow a^2 \geq 7 \).

Step 6: Find the Interval for \(a\)

Thus, \( a \) must satisfy \( a \leq -\sqrt{7} \) or \( a \geq \sqrt{7} \). Therefore, the interval is \(( -\infty, -\sqrt{7} ] \cup [ \sqrt{7}, \infty )\).

Key Concepts

Equation of a Circle

An equation of a circle in a coordinate plane is a mathematical representation of all the points that are a fixed distance, known as the radius, from a center point. The standard form of a circle's equation is written as \(x^2 + y^2 = r^2\), where \((x, y)\) are the coordinates of any point on the circle, and \(r\) is the radius. If the center is not at the origin, the general equation becomes \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle.
For instance, in our problem, the circle is centered at \((0, 0)\) with a radius of 5, giving the equation \(x^2 + y^2 = 25\). Understanding this concept is essential as it provides the geometric framework for analyzing relationships between shapes like circles and lines. Comprehending how these elements interact helps solve intersection problems, such as finding where a line intersects or is tangent to a circle.

Perpendicular Distance from Point to Line

To determine if the line intersects the circle at distinct points, coincident points, or does not intersect, we use the perpendicular distance from the center of the circle to the line. This distance provides insight into the relative position of the circle and the line. The formula for this distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}. \]
For our given circle, the center is at \((0, 0)\), and the equation of the line is \(3x + ay - 20 = 0\). When substituting, we find the perpendicular distance as \[ d = \frac{|0 - 20|}{\sqrt{3^2 + a^2}} = \frac{20}{\sqrt{9 + a^2}}. \]
This calculation helps us assess whether the line contacts the circle within its radius (meaning the line intersects the circle) or lies beyond it.

Inequality Solutions

Solving inequalities is crucial for determining the conditions under which the line intersects the circle. An inequality provides the range of values needed for a certain condition to hold true. In this context, we are interested in ensuring that the perpendicular distance \(d\) is not greater than the circle's radius (which is 5), leading to the inequality \[ \frac{20}{\sqrt{9 + a^2}} \leq 5. \]
By solving the inequality, we simplify it step by step: - Start by rearranging to \( \frac{20}{5} \leq \sqrt{9 + a^2} \),- This simplifies to \(4 \leq \sqrt{9 + a^2}\),- Squaring both sides results in \(16 \leq 9 + a^2\).
This further simplifies to \(a^2 \geq 7\). Hence, \(a\) must satisfy \( a \leq -\sqrt{7} \) or \( a \geq \sqrt{7} \), creating the interval \(( -\infty, -\sqrt{7} ] \, \cup \, [ \sqrt{7}, \infty )\). Thus, solving inequalities is essential in finding the range of values for which the line's interaction with the circle changes.