Problem 110 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The greatest distance of the point \(P(10,7)\) from the circle \(x^{2}+y^{2}-4 x-2 y-20=0\) is \(\quad\) (A) 10 unit (B) 15 unit (C) 5 unit (D) none of these

Step-by-Step Solution

Answer

The greatest distance is 15 units (option B).

Step 1: Identify Circle's Center and Radius

The given equation of the circle is \(x^2 + y^2 - 4x - 2y - 20 = 0\). We need to rewrite it in the standard form \((x-h)^2 + (y-k)^2 = r^2\). To do so, complete the square for both \(x\) and \(y\).

Step 2: Complete the Square for x

Start with the terms related to \(x\): \(x^2 - 4x\).Add and subtract \(4\) (since \((-4/2)^2 = 4\)) to get:\((x^2 - 4x + 4) - 4 = (x-2)^2 - 4\).

Step 3: Complete the Square for y

Next, complete the square for \(y\): \(y^2 - 2y\).Add and subtract \(1\) (since \((-2/2)^2 = 1\)) to get:\((y^2 - 2y + 1) - 1 = (y-1)^2 - 1\).

Step 4: Rewrite the Circle Equation

Substitute the completed squares back into the circle equation:\[(x-2)^2 - 4 + (y-1)^2 - 1 = 20\]\[(x-2)^2 + (y-1)^2 = 25\].Thus, the circle's center is \((2,1)\) and the radius is \(5\).

Step 5: Calculate Distance from Point P to the Circle's Center

Use the distance formula to find the distance from point \(P(10,7)\) to the center \((2,1)\) of the circle:\[\sqrt{(10-2)^2 + (7-1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 +36} = \sqrt{100} = 10\text{ units}\].

Step 6: Determine the Greatest Distance

The greatest distance from the point \(P\) to the circle is the sum of this distance to the center and the circle's radius:\[\text{Greatest distance} = 10 + 5 = 15 \text{ units}\].

Key Concepts

Circle Equation

In analytic geometry, the equation of a circle forms the basis for graphing or identifying attributes such as the center and radius of a circle. The standard form of a circle's equation is \[(x-h)^2 + (y-k)^2 = r^2\]Here,

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

If a circle's equation is given in another form, like the general quadratic form \(x^2 + y^2 + Dx + Ey + F = 0\), our task is to convert it to the standard form. This conversion often involves completing the square for both the \(x\) and \(y\) variables. Once a circle's equation is in standard form, it gives a clear view of the circle's center and radius, simplifying the analysis of its geometric properties.

Distance Formula

The distance formula is a critical tool in geometry for calculating the distance between two points in a plane. It is expressed as: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Where

• \((x_1, y_1)\) are the coordinates of the first point
• \((x_2, y_2)\) are the coordinates of the second point

This equation derives from the Pythagorean theorem, as it simply represents the hypotenuse of a right triangle formed by the differences in \(x\) and \(y\) coordinates.In our specific problem, the distance formula helps us find how far point \(P(10, 7)\) is from the circle's center \((2, 1)\). Recognizing this distance is crucial, as it forms part of the calculation for determining how far point \(P\) could be from any point on the circle, including the farthest possible point.

Completing the Square

Completing the square is a mathematical technique used primarily to transform a quadratic expression into a squared binomial, facilitating easier manipulation. This technique is especially useful in deriving the standard form of a circle's equation from its general form.For any quadratic expression like \[x^2 + bx\], we complete the square by adding and subtracting \[\left(\frac{b}{2}\right)^2\].This converts the expression into a perfect square trinomial, \[(x + \frac{b}{2})^2 - \left(\frac{b}{2}\right)^2\].In the example problem, completing the square is applied separately to the \(x\) terms and the \(y\) terms:

• \(x^2 - 4x\) becomes \((x-2)^2 - 4\)
• \(y^2 - 2y\) becomes \((y-1)^2 - 1\)

The adjustments allow us to rewrite the circle's equation, revealing both the circle's center and radius directly from the squared terms.