Problem 3 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the tangents \(P A\) and \(P B\) are drawn from the point \(P(-1,2)\) to the circle \(x^{2}+y^{2}+x-2 y-3=0\) and \(C\) is the centre of the circle, then the area of the quadrilateral \(P A C B\) is (A) 4 (B) 16 (C) does not exist (D) none of these

Step-by-Step Solution

Answer

The area of quadrilateral PACB is 4.

Step 1: Identify the Circle's Center and Radius

The equation of the circle is given as \(x^{2}+y^{2}+x-2y-3=0\). We first rewrite this equation in the standard form of a circle, \((x-h)^2 + (y-k)^2 = r^2\), by completing the square for both \(x\) and \(y\).\\[\begin{align*}\x^2 + x & : \text{Complete the square: } (x+\frac{1}{2})^2 - \frac{1}{4}\y^2 - 2y & : \text{Complete the square: } (y-1)^2 - 1\\end{align*}\] \Thus, the modified equation becomes: \\[(x + \frac{1}{2})^2 + (y - 1)^2 = \frac{15}{4}\] \The center \(C\) is \((-\frac{1}{2}, 1)\) and the radius \(r = \sqrt{\frac{15}{4}}\).

Step 2: Determine the Length of Tangents PA and PB

The length of the tangent from a point \(P(x_1, y_1)\) to a circle with center \((h, k)\) and radius \(r\) can be calculated using the formula \(\sqrt{x_1^2+y_1^2-2hx_1-2ky_1+c}\) where the circle equation is \(x^2 + y^2 + 2gx + 2fy + c = 0\). \Substituting, for \(P(-1,2)\), we compute as follows: \[\text{Length} = \sqrt{(-1)^2 + (2)^2 - 2\left(-\frac{1}{2}\right) - 2(1) + (-3)} = \sqrt{4}\]. \Thus, \(PA = PB = 2\).

Step 3: Compute Distance from P to C

We need the distance \(PC\) where \(P\) is \((-1,2)\) and \(C\) is \((-\frac{1}{2}, 1)\). Use the distance formula \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\). \[\begin{align*}\text{Distance } PC & = \sqrt{\left(-\frac{1}{2} - (-1)\right)^2 + (1 - 2)^2} \& = \sqrt{\left(\frac{1}{2}\right)^2 + (-1)^2} \& = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \end{align*}\]

Step 4: Area of Quadrilateral PACB

The area of a quadrilateral formed by two tangents and a line from an external point to the circle's center can be calculated from triangle properties. In this case, notice \(PACB\) forms a kite with two right triangles \(\triangle PAC\) and \(\triangle PBC\) sharing a base \(PC\). The area of \(PACB\) is calculated by \(\text{Area}(PACB) = PA \times PB\) since \(PA = PB\) and both \(PA\) and \(PB\) are tangents, making right angles with \(PC\).Thus, \(\text{Area} = PA \times PB = 2 \times 2 = 4\).

Key Concepts

Tangents to a Circle

In geometry, a tangent to a circle is a straight line that touches the circle at exactly one point. This point is known as the point of tangency. Understanding tangents is crucial because they directly relate to other geometric properties and theorems.
The key characteristics of tangents include:

• A tangent never crosses a circle; it only touches at one point.
• The radius of the circle at the point of tangency is perpendicular to the tangent.
• The length of tangents from a single external point to where they touch a circle is equal.

For the problem above, two tangents, denoted as \(PA\) and \(PB\), are drawn from a point \(P(-1, 2)\) to a circle. We found that these tangents are equal in length due to the property of tangents drawn from an external point. This makes solving geometry problems involving circles and tangents both systematic and predictable.

Circle Equation

The circle equation in its standard form is oindent \((x-h)^2 + (y-k)^2 = r^2,\)where:

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

Converting a given circle equation into this form involves completing the square for both \(x\) and \(y\). This allows us to easily identify the circle's center and radius.
In the original problem, we started with the equation:\[x^{2} + y^{2} + x - 2y - 3 = 0\] Completing the square gives us: \[(x + \frac{1}{2})^2 + (y - 1)^2 = \frac{15}{4}.\] From this, it's evident that the center \(C\) is \((-\frac{1}{2}, 1)\) and the radius \(r\) is \(\sqrt{\frac{15}{4}}\). Recognizing how this transformation works helps in various geometry and calculus applications, as identifying these elements is essential for further calculations.

Distance Formula

The distance formula is an essential geometric formula used to determine the distance between two points in a coordinate plane. This formula is derived from the Pythagorean theorem and is expressed as:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Here,

• \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.
• \(d\) is the distance between these two points.

In the exercise provided, the distance formula was applied to find the distance \(PC\), where \(P(-1,2)\) is the external point, and \(C(-\frac{1}{2}, 1)\) is the center of the circle. Calculating this gives us:\[PC = \sqrt{\left(\frac{1}{2}\right)^2 + (-1)^2} = \frac{\sqrt{5}}{2}.\]Understanding the distance formula is critical as it is used in various fields including physics, engineering, and navigation, determining not only lengths but also relationships between points in space.