Problem 111 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation of the tangent to the circle \(x^{2}+y^{2}+4 x-\) \(4 y+4=0\) which make equal intercepts on the positive co-ordinate axes, is (A) \(x+y=2\) (B) \(x+y=2 \sqrt{2}\) (C) \(x+y=4\) (D) \(x+y=8\)

Step-by-Step Solution

Answer

The equation of the tangent line is \(x + y = 2\sqrt{2}\) (Option B).

Step 1: Identify the form of the equation

Recognize that the equation of the circle is in the form \(x^2 + y^2 + 2gx + 2fy + c = 0\). Here, it's \(x^2 + y^2 + 4x - 4y + 4 = 0\). This implies \(2g = 4\), \(2f = -4\), and \(c = 4\). Hence, the center of the circle \((h, k)\) is given by \((-g, -f) = (-2, 2)\).

Step 2: Calculate the radius of the circle

Use the formula for the radius \(r\) of the circle: \(r = \sqrt{g^2 + f^2 - c}\). Here, \(g = 2\), \(f = -2\), and \(c = 4\). Calculate \(r = \sqrt{2^2 + (-2)^2 - 4} = \sqrt{4 + 4 - 4} = \sqrt{4} = 2\).

Step 3: Determine the equation of the tangent line

The tangent line makes equal intercepts on the positive coordinate axes, meaning it is of the form \(x+y=c\) where \(c\) determines the length of these intercepts. Here, the tangent line must be tangent to the circle.

Step 4: Evaluate the specific tangent condition

For a line \(x+y=c\) to be tangent to a circle with center \((-2,2)\) and radius 2, the perpendicular distance \(d\) from the center to the line must equal the radius \(r\). This distance is \(d = \frac{|(-2)+2-c|}{\sqrt{1^2 + 1^2}} = \frac{|c|}{\sqrt{2}}\). Set \(\frac{|c|}{\sqrt{2}} = 2\).

Step 5: Solve for the intercept \(c\)

Solve \(\frac{|c|}{\sqrt{2}} = 2\). Multiplying both sides by \(\sqrt{2}\) gives \(|c| = 2\sqrt{2}\). Thus, \(c = 2\sqrt{2}\).

Step 6: Confirm the correct equation

Given the options, the correct equation is the line \(x + y = 2\sqrt{2}\), which achieves the condition of making equal positive intercepts and being tangent to the circle.

Key Concepts

Equation of a Circle

The equation of a circle in a two-dimensional coordinate plane is a way to describe all the points that are equidistant from a fixed center point. The general form of the equation of a circle is given by:\[x^2 + y^2 + 2gx + 2fy + c = 0\]- The coefficients \(2g\) and \(2f\) help to locate the center of the circle at \((-g, -f)\).- The constant term \(c\) helps in determining the size of the circle, specifically in calculating the radius.The center \((h, k)\) and radius \(r\) of the circle can be derived using the relationships:\[h = -g, \quad k = -f\]\[r = \sqrt{g^2 + f^2 - c}\]In our exercise, breaking down the given equation \(x^2 + y^2 + 4x - 4y + 4 = 0\), we found the center to be \((-2, 2)\) and the radius to be 2. Understanding how to transform this standard form to determine the circle's geometric properties is crucial for solving problems related to circle geometry.

Intercepts of a Line

The intercepts of a line are points at which the line crosses the coordinate axes. To find the intercepts of a line, identify where the line meets the x-axis and y-axis.When dealing with a line equation in the form \(x + y = c\), both the x-intercept and the y-intercept can be determined.- **x-intercept**: Set \(y = 0\) and solve for \(x\).- **y-intercept**: Set \(x = 0\) and solve for \(y\).For the line to make equal intercepts on both axes, this means each intercept is the same distance from the origin. Thus, the intercepts are given by \(x = c\) and \(y = c\). In this exercise, we're looking at lines of the form \(x+y=c\), ensuring both intercepts are equal. To check tangency with a circle, the perpendicular distance from the circle's center to this line must be determined.

Circle Geometry

Circle geometry covers the properties and relations of lines and points in relation to a circle. One crucial aspect is understanding tangents. A tangent is a straight line that touches the circle at exactly one point and does not cross it.Here's a brief overview of tangents when dealing with circle geometry:- **Single point contact**: A tangent to a circle will touch the circle at only one point. This point is called the point of tangency.- **Perpendicular radius**: The radius that extends to the point of tangency is perpendicular to the tangent line.For a line to be tangent to a circle, the distance from the line to the circle's center must equal the radius of the circle. In this exercise, for the tangent line represented by \(x + y = 2\sqrt{2}\), the perpendicular distance from the center \((-2, 2)\) to this line equalled the circle's radius, verifying the tangent condition.Knowing these relationships aides substantially in solving geometric problems involving circles and their tangents.