Problem 47 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The common chord of the circle \(x^{2}+y^{2}+8 x+4 y-5=\) 0 and a circle passing through the origin and touching the line \(y=x\), passes through the fixed point (A) \(\left(\frac{5}{12}, \frac{5}{12}\right)\) (B) \(\left(\frac{5}{12}, \frac{-5}{12}\right)\) (C) \(\left(\frac{-5}{12}, \frac{5}{12}\right)\) (D) none of these

Step-by-Step Solution

Answer

The fixed point is none of these options.

Step 1: Identify the Equation of First Circle

The given circle equation is \(x^{2}+y^{2}+8x+4y-5=0\). To find its center and radius, rewrite it in center-radius form:Complete the square for both \(x\) and \(y\):\[ (x^2 + 8x) + (y^2 + 4y) = 5 \]Complete square \((x^2 + 8x + 16) + (y^2 + 4y + 4) = 5 + 16 + 4 \)This simplifies to:\((x+4)^2 + (y+2)^2 = 25 \)Thus, the circle has center \((-4, -2)\) and radius 5.

Step 2: Define the Second Circle

The second circle passes through the origin and touches the line \(y=x\). If the equation of this circle is assumed as \(x^2 + y^2 + 2gx + 2fy + c = 0\) and it passes through the origin (0,0), we have:\[ c = 0 \]Thus, the equation becomes \(x^2 + y^2 + 2gx + 2fy = 0\). Since it touches the line \(y = x\), the perpendicular distance from the center \((-g, -f)\) to the line \(y = x\) should be equal to the radius of the circle.

Step 3: Perpendicular Distance Condition

The perpendicular distance from \((-g, -f)\) to the line \(y = x\) is given by:\[\frac{|-g + f|}{\sqrt{2}}\]This distance should be equal to the radius of the circle (let's denote it \(r\)). Therefore:\[\frac{|-g + f|}{\sqrt{2}} = r\]

Step 4: Find Common Chord Equation

The equation of the common chord between two circles \(S_1 = 0\) and \(S_2 = 0\) is \(S_1 - S_2 = 0\).The first circle (\(S_1\)) is \(x^2+y^2+8x+4y-5=0\), and for the second circle \(S_2 = x^2 + y^2 + 2gx + 2fy = 0\), it simplifies to:\[ 6x + 2y - 5 - 2gx - 2fy = 0 \]Simplifying, we obtain:\[ (6 - 2g)x + (2 - 2f)y = 5 \]

Step 5: Check for Fixed Point

The line from Step 4 can be written as:\[ (3 - g)x + (1 - f)y = \frac{5}{2} \]For the line to pass through a fixed point independent of \(g\) and \(f\), substitute the provided options into the equation to check which of them satisfy it regardless of \(g\) and \(f\). By substitution and elimination process, we find that no value of \((x, y)\) satisfies the equation independent of \(g\) and \(f\), pointing to none of the specific options.

Key Concepts

Circle Equation

Understanding a circle equation is essential in coordinate geometry. A circle can be defined by an equation in the form

• Standard form: \[(x - h)^2 + (y - k)^2 = r^2 \]where \((h, k)\) is the center of the circle, and \(r\) is the radius.
• General form: \[x^2 + y^2 + Dx + Ey + F = 0 \] This form might initially look complicated, but with some algebraic manipulation, such as completing the square, it can be transformed into the standard form.

These equations help determine the properties of the circle, like its center and radius. For instance, turning the general form into the standard form reveals the circle's center coordinates and its radius — crucial characteristics when analysing geometrical patterns and their relationships.
In the case presented, starting from the given circle equation allows us to identify and manipulate these inherent properties effectively.

Perpendicular Distance

In geometry, perpendicular distance between a point and a line is integral in various applications, including determining tangency — as seen in this problem. The line here is expressed in the form

• Linear: \[y = x \]

When calculating the perpendicular distance

• Use the formula: \[ \text{Distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]where \(a\), \(b\), and \(c\) are coefficients from the line's equation and \((x_1, y_1)\) is a point.
• Here, the perpendicular distance is derived specifically from \((-g, -f)\) to \(y=x\).

The significance of using this approach is that it reveals when a circle touches a line. When this perpendicular distance equals the radius, it confirms tangency between the circle and the line, a key step in solving these types of geometrical problems.

Center-Radius Form

Transforming a general circle equation into its center-radius form, we harness completing the square technique to pinpoint any circle's physical properties. In such transformations, start with

• The general circle equation: \[x^2 + y^2 + Dx + Ey + F = 0 \]
• Completion involves rearranging terms to complete perfect squares: \[(x + h)^2 + (y + k)^2 = r^2 \]Here, \(-h\) and \(-k\) become the transformed center's coordinates.

For instance, reviewing the derived equation from the given circle, the center is \((-4, -2)\), with a radius of \(5\).
This form offers simplicity and clarity, making computations and deductions more straightforward, particularly crucial when working with complex problems involving interactions with lines or other circles.

Line Equation

Grasping the concept of a line equation, especially in its various forms, equips you with the tools to solve intricate geometry problems. Lines in the plane can be represented as

• Standard linear: \[y = mx + c \]where \(m\) is the slope and \(c\) the y-intercept.
• General form: \[Ax + By + C = 0\]which is versatile and can accommodate any line position within the Cartesian plane.

The process of finding the common chord, which is the radical axis between two intersecting circles, entails combining the two equations of the circles and simplifying them into a line equation. This chord is ultimately the set intersection points, forming a straight line.
Recognizing different equations and manipulating them to find a solution is crucial, as illustrated in this exercise where substitution and comparison yield insight into the intricacies of circle and line relationships.