Problem 60 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The point on the straight line \(y=2 x+11\) which is nearest to the circle \(16\left(x^{2}+y^{2}\right)+32 x-8 y-50=0\) is (A) \(\left(\frac{9}{2}, 2\right)\) (B) \(\left(-\frac{9}{2}, 2\right)\) (C) \(\left(\frac{9}{2},-2\right)\) (D) none of these

Step-by-Step Solution

Answer

(B) \((-\frac{9}{2}, 2)\)

Step 1: Rewrite the Circle Equation

The given circle equation is \(16(x^2 + y^2) + 32x - 8y - 50 = 0\). First, let's rewrite this equation by rearranging and simplifying using algebraic methods. Divide the entire equation by 16 to simplify it:\[x^2 + y^2 + 2x - \frac{1}{2}y - \frac{25}{8} = 0\]

Step 2: Complete the Square

Next, complete the square for the terms involving \(x\) and \(y\) to convert the circle equation into standard form. Starting with \(x\) terms:- Take the term \(x^2 + 2x\). - To complete the square, add and subtract \((\frac{2}{2})^2 = 1\): \[x^2 + 2x = (x+1)^2 - 1\]Similarly, for \(y\) terms, take \(y^2 - \frac{1}{2}y\):- Add and subtract \((\frac{-1/2}{2})^2 = \frac{1}{16}\): \[y^2 - \frac{1}{2}y = (y - \frac{1}{4})^2 - \frac{1}{16}\]

Step 3: Form the Circle's Standard Equation

Substitute the completed square terms back into the equation and rearrange:\[(x+1)^2 - 1 + (y - \frac{1}{4})^2 - \frac{1}{16} - \frac{25}{8} = 0\]Combine constants to simplify:\[(x+1)^2 + (y - \frac{1}{4})^2 = \frac{81}{16}\]This is now a standard form of a circle equation with center \((-1, \frac{1}{4})\) and radius \(\frac{9}{4}\).

Step 4: Express Line in Parametric Form

The line is \(y = 2x + 11\). We can express a point \((x, y)\) on the line in terms of \(x\) alone:- Let \(x = t\). Then \(y = 2t + 11\).- So, a point on the line can be parameterized as \((t, 2t + 11)\).

Step 5: Set Up Distance Formula

The task is to minimize the distance between a point \((t, 2t + 11)\) on the line and the circle's center \((-1, \frac{1}{4})\).The distance \(d\) between these two points is given by:\[d = \sqrt{(t + 1)^2 + (2t + 11 - \frac{1}{4})^2}\]

Step 6: Calculate and Minimize Squared Distance

Minimizing \(d\) is the same as minimizing \(d^2\). Let's calculate \(d^2\):\[d^2 = (t + 1)^2 + (2t + \frac{43}{4})^2\]Expand and differentiate this expression with respect to \(t\) to find the minimum:

Step 7: Differentiation and Solve for Critical Point

Expand \(d^2 = (t+1)^2 + (2t + \frac{43}{4})^2\), differentiate with respect to \(t\) and set the derivative to zero to find critical points. Then solve:\[\frac{d}{dt}\left((t+1)^2 + (2t+\frac{43}{4})^2\right) = 2(t+1) + 4(2t + \frac{43}{4}) = 0 \]Simplify and solve this linear equation to find \(t\).

Step 8: Evaluate t and Find Nearest Point

Solve the equation from step 7:- Simplifying gives: \[2t + 2 + 8t + \frac{43}{2} = 0\] \[10t + \frac{47}{2} = 0\] \[10t = -\frac{47}{2}\] \[t = -\frac{47}{20}\]Substitute this back in the line equation \((t, 2t+11)\) for \(t\), and calculate: \[x = -\frac{47}{20}, \quad y = 2(-\frac{47}{20}) + 11 = 2\cdot (-\frac{47}{20}) + \frac{220}{20} = \frac{9}{2}\]Thus, the point is \(\left(-\frac{9}{2}, 2\right)\).

Step 9: Conclusion

Based on calculations, the point on the line \(y = 2x + 11\) closest to the circle \((x+1)^2 + (y - \frac{1}{4})^2 = \frac{81}{16}\) is \((-\frac{9}{2}, 2)\), corresponding to option (B).

Key Concepts

Circle Equation

The circle equation is a fundamental concept in coordinate geometry. It helps us to represent a circle on the Cartesian plane using an algebraic expression. The general form is \[Ax^2 + By^2 + Cx + Dy + E = 0\]which can be transformed into the standard form. The standard form, \[(x - h)^2 + (y - k)^2 = r^2\], is more insightful:

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

This transformation involves simplifying and completing the square, essential techniques in coordinate geometry.
Understanding the different forms of the circle equation assists in solving various geometrical problems, like locating points or measuring distances.

Parametric Form

The parametric form is a way of expressing the coordinates of points on geometric shapes using a parameter. In this context, converting the line equation \(y = 2x + 11\) into parametric form simplifies finding specific points on the line.

• Choose a parameter, such as \(t\) for \(x\), leading to \(x = t\).
• Substitute into the line equation: \(y = 2t + 11\).
• The line can then be expressed as points \((t, 2t+11)\).

This form is particularly useful when dealing with intersections or minimizing distances.

Distance Formula

The distance formula provides a way to calculate the distance between two points in a plane. Given two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) is expressed as: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]In problems involving geometry, such as finding the closest point on a line to a circle, this formula is key.

• Calculate the difference in \(x\) and \(y\) coordinates.
• Square these differences, sum them, and take the square root.

Ironically, in optimization problems, minimizing \(d^2\) instead of \(d\) can simplify calculations. This simplifies avoiding the square root operation.

Completing the Square

Completing the square is a method used to transform quadratics into a perfect square trinomial. It is vital when converting the general circle equation into standard form. Here's the process:

• Identify the quadratic terms in \(x\) and \(y\).
• For each, add and subtract the square of half the linear coefficient to form a perfect square.
• This reduces expressions like \(x^2 + bx\) into \((x + \frac{b}{2})^2 - (\frac{b}{2})^2\).

This technique reveals the center and the radius of the circle. Completing the square balances the equation and makes it easier to work with geometrically.