Problem 102 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

I. The tangent to the cir- (A) \(\left(-\frac{3}{2}, \frac{1}{2}\right)\) cle \(x^{2}+y^{2}=9\), which is parallel to \(y\)-axis and does not lie in third quadrant, touches the circle at the point II. The coordinates of the (B) \((1,1)\) middle point of the chord which the circle \(x^{2}+y^{2}+4 x-2 y-3=\) 0 cuts off on the line \(y\) \(=x+2\), are III. The circle passing (C) \((3,0)\) through three distinct points \((1, t),(t, 1)\) and \((t)\) \(t\) ) passes through the point IV. The chords of contact (D) \(\left(\frac{1}{2}, \frac{1}{4}\right)\) of the pair of tangents drawn from each point on the line \(2 x+y=4\) to the circle \(x^{2}+y^{2}=1\) pass through the fixed point

Step-by-Step Solution

Answer

I: (3,0); II: (1,1); III: (3,0); IV: (-3/2, 1/2).

Step 1: Understanding the Tangent Problem

We are given a circle with the equation \(x^2 + y^2 = 9\), which is a circle centered at the origin with a radius of 3. A tangent line to this circle is parallel to the y-axis, which means it is a vertical line. The equation of a vertical line is in the form \(x = a\). Since it does not lie in the third quadrant, \(a\) must be non-negative. The distance from the center to the tangent \(|a|\) must be equal to the radius, thus \(a = \pm 3\). Because it doesn't lie in the third quadrant, the coordinate is \((3,y)\). The point of tangency is \((3,0)\), since in third quadrant both coordinates would otherwise be negative.

Step 2: Middle Point of a Chord on a Given Line

The equation of the circle is \(x^2 + y^2 + 4x - 2y - 3 = 0\), which we can rewrite as \((x+2)^2 + (y-1)^2 = 8\) by completing the square. The equation of the line is \(y = x + 2\). Substitute \(y = x + 2\) into the circle equation and solve for \(x\). After finding the points where the line intersects the circle, we calculate the midpoint of the chord. Solving gives the midpoint \((1,1)\).

Step 3: Finding a Common Point for Circle through Given Points

The circle passes through the points \((1,t)\), \((t,1)\), and \((t,t)\). To find a common point that it passes through, consider the symmetry of these points. By substituting these points into the general form of the equation of a circle and using properties of symmetric points, the circle must pass through \((3,0)\).

Step 4: Chords of Contact and Fixed Point Analysis

The circle is \(x^2 + y^2 = 1\), and we use the equation of the chord of contact from a point \((x_1,y_1)\) as \(xx_1 + yy_1 = 1\). Simplifying with the given line \(2x + y = 4\), substitute points from the line into the chord equation to find the fixed point through which all such chords pass, resulting in \(\left(-\frac{3}{2}, \frac{1}{2}\right)\).

Key Concepts

Exploring Geometry Problems

Geometry problems often involve spatial reasoning and understanding the properties and relationships of geometric figures. Tangents are lines that just touch a circle at one point.
In the given exercise, we consider a circle centered at the origin and a vertical tangent parallel to the y-axis. This setup relies on understanding that vertical tangents have the form \(x = a\), where \(a\) is a constant.
The problem requires calculating the tangential point of contact that lies outside the third quadrant. Here, knowing that the radius of the circle is 3 aids in finding that the line \(x = 3\) is vertical and valid since it does not lie in the third quadrant. The tangential touchpoint thus lies at \((3, 0)\).

• Vertical tangents imply a constant x-coordinate.
• Tangent must ensure a single point of touch with the circle.
• The circle's radius determines the possible position of tangents.

Deciphering Equations of Circles

The equation \(x^2 + y^2 = 9\) represents a classic circle centered at the origin with a radius of 3. Such equations provide precise locations and boundaries of the circle.
Another form, \((x+2)^2 + (y-1)^2 = 8\), illustrates a circle not centered at the origin but at \((-2, 1)\) with \(\sqrt{8}\) as its radius.
By rewriting the equation of a circle, for instance, using the method of completing the square, one can derive the circle's center and radius directly from the equation.

• Standard form: \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) the radius.
• Completing the square helps in converting to standard form.
• Understanding the equation helps in geometric interpretations and problem-solving.

Navigating Coordinate Geometry

Coordinate geometry marries algebra and geometry, allowing for precise derivation and analytical study of geometric figures. It offers a visual understanding through equations describing curves and their properties.
In problems concerning circles and lines, as in the exercise, making line equations \(y = x + 2\) work with circle equations like \(x^2 + y^2 + 4x - 2y - 3 = 0\) requires substitutions to find intersection points representing chords. These visible intersections allow one to calculate midpoints directly.
The line-chord relationship showcases coordinate geometry's power, highlighting how coordinates amalgamate the equation-based solutions with graphical understanding for geometry problems.

• Combines algebraic equations with geometric figures.
• Provides method to solve for coordinates using equations.
• Useful for modeling and analyzing intersections, tangents.