Problem 5 - A Complete Resource Book in Mathematics for JEE Main

Question

The number of points on the circle \(x^{2}+y^{2}-4 x-10 y+\) \(13=0\) which are at a distance 1 from the point \((-3,2)\) is (A) \(\underline{1}\) (B) 2 (C) 3 (D) none of these

Step-by-Step Solution

Answer

(D) none of these

Step 1: Rewrite the circle equation in standard form

The given equation of the circle is \( x^2 + y^2 - 4x - 10y + 13 = 0 \). To rewrite it in standard form, we need to complete the square for both \( x \) and \( y \).For \( x \): - Take the coefficient of \( x \), which is \(-4\), divide it by 2 to obtain \(-2\), and square it to get \(4\).For \( y \):- Take the coefficient of \( y \), which is \(-10\), divide it by 2 to obtain \(-5\), and square it to get \(25\).The equation becomes:\[(x - 2)^2 + (y - 5)^2 = 4 + 25 - 13 = 16\]This is the standard form: \((x - 2)^2 + (y - 5)^2 = 16\) with center \((2, 5)\) and radius 4.

Step 2: Distance from point to center of circle

Calculate the distance between the center of the circle \((2, 5)\) and the point \((-3, 2)\). The distance formula is:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substituting \((x_1, y_1) = (2, 5)\) and \((x_2, y_2) = (-3, 2)\), we get:\[ d = \sqrt{((-3) - 2)^2 + (2 - 5)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}\]

Step 3: Determine known distances and check conditions

The radius of the circle is 4 (from the circle's standard equation). We want points that are 1 unit away from point \((-3, 2)\).First, check if \(\sqrt{34} + 1 = 4 \) or \( |\sqrt{34} - 1| = 4 \). These conditions determine whether a point can lie on the circle given its distance from \((-3, 2)\).Calculate:- \( \sqrt{34} + 1 \approx 5.83 \), not equal to 4.- \( |\sqrt{34} - 1| = \sqrt{34} - 1 \approx 4.83 \), not equal to 4.Neither condition holds.

Step 4: Conclude lack of intersecting points

Since neither condition from Step 3 results in an intersection of the circle with a point exactly 1 unit away from \((-3, 2)\), there are no points that satisfy all conditions of the problem.

Key Concepts

Standard form of circle

The standard form of a circle's equation is a crucial concept in understanding circles on the Cartesian coordinate system. A circle's equation in standard form is written as \[ (x - h)^2 + (y - k)^2 = r^2 \]where

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

To convert a given circle equation into this standard form, we often need to complete the square for both the \(x\) and \(y\) terms. We focus on simplifying to identify the center and radius succinctly. This process helps us visualize the circle more effectively on a graph by easily identifying its center and radius.

Distance formula

The distance formula is a fundamental tool used to measure the space between two points on a plane. When given two points,\((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) between these points can be calculated using the formula:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]This formula is derived from the Pythagorean theorem and allows us to calculate the straight-line distance between any two points. In our example, this formula helps us determine how far the point \((-3, 2)\) is from the center of the circle \((2, 5)\). Knowing how to calculate this distance is essential in many geometrical problems and can help determine relative positions, as well as solve for intersections and other related calculations.

Completing the square

Completing the square is a vital algebraic technique used to transform a quadratic equation into a perfect square trinomial. This process is especially useful when working with circle equations, where it assists in reformatting them into the standard form. To complete the square, you:

• Identify the coefficient of the linear term (e.g., \(x\) or \(y\)).
• Divide it by 2, then square the result.
• Add and subtract this square number within the equation to maintain balance.

For instance, in the circle equation\(x^2 + y^2 - 4x - 10y + 13 = 0\), we complete the square for \(x\) and \(y\), leading to the form \((x - 2)^2 + (y - 5)^2 = 16\). This technique helps in simplifying the equation for easier understanding and graphical representation.

Intersections of geometric shapes

Understanding the intersections of geometric shapes is crucial in determining where and how these shapes meet or intersect each other. In geometry, questions often ask us to find points, lines, or shapes that intersect or share common space with circles. In the context of our circle problem, we're trying to find points that are 1 unit away from \((-3, 2)\) and lie on the circle defined by \((x - 2)^2 + (y - 5)^2 = 16\). This requires checking additional conditions using distances, such as those derived from the distance formula, against the circle's radius.If the conditions don't match known distances or constraints, such as the circle's radius, it implies no intersection. Knowing how to check these conditions allows us to determine possibilities for such intersections in complex geometrical arrangements.