Problem 23 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The point \((1,4)\) lies inside the circle \(x^{2}+y^{2}-6 x-10 y\) \(+p=0\) which does not touch or intersect the coordinate axes, then (A) \(0

Step-by-Step Solution

Answer

(D) \(9 < p < 29\)

Step 1: Express the Circle in Standard Form

The equation of the circle is \(x^{2}+y^{2}-6x-10y+p=0\). To express it in standard form, complete the square for both \(x\) and \(y\).

Step 2: Complete the Square for x

For the \(x\)-terms: \(x^2 - 6x\), add and subtract \((6/2)^2 = 9\). This gives us \((x-3)^2 - 9\).

Step 3: Complete the Square for y

For the \(y\)-terms: \(y^2 - 10y\), add and subtract \((10/2)^2 = 25\). This gives us \((y-5)^2 - 25\).

Step 4: Rewrite the Standard Circle Equation

Substitute the completed squares back into the equation: \((x-3)^2 - 9 + (y-5)^2 - 25 + p = 0\). Simplify to find \((x-3)^2 + (y-5)^2 = 34 - p\).

Step 5: Inequality for the Point Inside the Circle

The distance from the center \((3, 5)\) to the point \((1, 4)\) should be less than the radius. Calculate the distance: \( \sqrt{(3-1)^2 + (5-4)^2} = \sqrt{5} \). The inequality is then \( \sqrt{5} < \sqrt{34-p} \).

Step 6: Solve the Inequality for p

Square both sides of the inequality: \(5 < 34 - p\). Simplify to find \(p < 29\).

Step 7: Consider Circle Not Touching Coordinate Axes

The circle should not touch the coordinate axes, implying \((3-R > 0)\) and \((3+R < 0)\), where \(R^2 = 34-p\). Thus, both conditions provide a range for \(R\), but it's given the circle does not intersect, so we check \((x-R)^2 + (y + R)^2 < 29\).

Step 8: Combine Conditions

From steps 6 and 7, it requires that \(9

Key Concepts

Completing The Square

Completing the square is a method used to rewrite quadratic expressions in a form that reveals important properties of a function, such as its vertex or center in geometry problems. This technique is often used to convert equations into a form that makes them easier to solve or analyze.

In the case of circle equations, completing the square helps us transform the equation into a standard form, revealing the circle's center and radius. Consider the quadratic term that involves a variable, for example: given the expression \(x^2 - 6x\), we aim to turn it into a perfect square trinomial. To complete the square, we use the following steps:

• Take the coefficient of the linear term, divide it by two, and square it. For \(x^2 - 6x\), we calculate \((-6/2)^2 = 9\).
• Add and subtract this number within the expression to maintain its balance forming \((x-3)^2 - 9\).

This process is repeated for both x and y terms, allowing the circle equation to take on a familiar form.

Standard Form of a Circle

The standard form of a circle is derived by rewriting a general circle equation like \(x^2 + y^2 - 6x - 10y + p = 0\) in a form that directly shows the circle's center and radius.

This form can be expressed as:\[ (x-h)^2 + (y-k)^2 = r^2 \]
where \((h, k)\) is the center of the circle, and \(r\) is the radius.

From the previous exercise, we completed the square to convert \(x^2 - 6x\) and \(y^2 - 10y\) into \((x-3)^2 - 9\) and \((y-5)^2 - 25\), respectively. Substituting these back provides us with:
\[ (x-3)^2 + (y-5)^2 = 34 - p \]
The transformed expression clearly shows that the circle is centered at \((3,5)\) and its radius squared is \(34 - p\). This allows us to more easily work with the circle, understand its properties, and utilize it when solving geometry problems.

Inequalities

In mathematics, inequalities are expressions involving the symbols \(<, \leq, >, \geq\) that show the relationship between two values that are not equal. They are crucial in defining regions and in determining conditions for various geometric shapes.

For circles, inequalities help define where points lie concerning the circle:

• Points inside the circle satisfy \((x-h)^2 + (y-k)^2 < r^2\).
• Points on the circle satisfy \((x-h)^2 + (y-k)^2 = r^2\).
• Points outside the circle satisfy \((x-h)^2 + (y-k)^2 > r^2\).

For our specific example, the point \((1,4)\) is within the circle centered at \((3,5)\) with radius squared of \(34 - p\). The condition \(\sqrt{5} < \sqrt{34-p}\) is derived from calculating the distance from the circle center to \((1,4)\), giving a range for \(p\) when solved, indicating \(p < 29\). This approach ensures that the specified point remains inside the circle.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves using algebra and the Cartesian coordinate system to explore geometric properties and relationships.

This field allows us to analyze shapes, lines, and curves by applying coordinate-based methods. For circles, which are a fundamental shape in geometry, coordinate geometry helps in understanding and solving problems through equations.

The standard circle equation \((x-3)^2 + (y-5)^2 = 34 - p\) is set in the Cartesian plane. The center of the circle at \((3,5)\) is vital for determining distances and relationships with other points and lines in the plane. Coordinate geometry aids in visualizing and solving problems relating the circle to axes. For example, ensuring the circle does not touch the axes involves understanding the circle's constraints and positioning. It's essential to know how the radius and center relate to each given point or line within the coordinate plane to solve intersecting problems effectively.

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Problem 22

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Problem 24

Other exercises in this chapter

Problem 21

If the locus of a point which moves so that the line joining the points of contact of the tangents drawn from it to the circle \(x^{2}+y^{2}=b^{2}\) touches the

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Problem 22

A variable circle passes through the fixed point \(A(p, q)\) and touches \(x\)-axis. The locus of the other end of the diameter through \(A\) is (A) \((y-q)^{2}

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Problem 24

A circle \(C_{1}\) of radius 2 touches both \(x\)-axis and \(y\)-axis. Another circle \(C_{2}\) whose radius is greater than 2 touches circle \(C_{1}\) and both

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Problem 25

The equation to the sides \(A B, B C, C A\) of a \(\triangle A B C\) are \(x+y=1,4 x-y+4=0\) and \(2 x+3 y=6 .\) Circles are drawn on \(A B, B C, C A\) as diame

View solution

Key Concepts

Completing The Square

Completing the square is a method used to rewrite quadratic expressions in a form that reveals important properties of a function, such as its vertex or center in geometry problems. This technique is often used to convert equations into a form that makes them easier to solve or analyze.

In the case of circle equations, completing the square helps us transform the equation into a standard form, revealing the circle's center and radius. Consider the quadratic term that involves a variable, for example: given the expression \(x^2 - 6x\), we aim to turn it into a perfect square trinomial. To complete the square, we use the following steps:

• Take the coefficient of the linear term, divide it by two, and square it. For \(x^2 - 6x\), we calculate \((-6/2)^2 = 9\).
• Add and subtract this number within the expression to maintain its balance forming \((x-3)^2 - 9\).

This process is repeated for both x and y terms, allowing the circle equation to take on a familiar form.

Standard Form of a Circle

The standard form of a circle is derived by rewriting a general circle equation like \(x^2 + y^2 - 6x - 10y + p = 0\) in a form that directly shows the circle's center and radius.

This form can be expressed as:\[ (x-h)^2 + (y-k)^2 = r^2 \]
where \((h, k)\) is the center of the circle, and \(r\) is the radius.

From the previous exercise, we completed the square to convert \(x^2 - 6x\) and \(y^2 - 10y\) into \((x-3)^2 - 9\) and \((y-5)^2 - 25\), respectively. Substituting these back provides us with:
\[ (x-3)^2 + (y-5)^2 = 34 - p \]
The transformed expression clearly shows that the circle is centered at \((3,5)\) and its radius squared is \(34 - p\). This allows us to more easily work with the circle, understand its properties, and utilize it when solving geometry problems.

Inequalities

In mathematics, inequalities are expressions involving the symbols \(<, \leq, >, \geq\) that show the relationship between two values that are not equal. They are crucial in defining regions and in determining conditions for various geometric shapes.

For circles, inequalities help define where points lie concerning the circle:

• Points inside the circle satisfy \((x-h)^2 + (y-k)^2 < r^2\).
• Points on the circle satisfy \((x-h)^2 + (y-k)^2 = r^2\).
• Points outside the circle satisfy \((x-h)^2 + (y-k)^2 > r^2\).

For our specific example, the point \((1,4)\) is within the circle centered at \((3,5)\) with radius squared of \(34 - p\). The condition \(\sqrt{5} < \sqrt{34-p}\) is derived from calculating the distance from the circle center to \((1,4)\), giving a range for \(p\) when solved, indicating \(p < 29\). This approach ensures that the specified point remains inside the circle.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves using algebra and the Cartesian coordinate system to explore geometric properties and relationships.

This field allows us to analyze shapes, lines, and curves by applying coordinate-based methods. For circles, which are a fundamental shape in geometry, coordinate geometry helps in understanding and solving problems through equations.

The standard circle equation \((x-3)^2 + (y-5)^2 = 34 - p\) is set in the Cartesian plane. The center of the circle at \((3,5)\) is vital for determining distances and relationships with other points and lines in the plane. Coordinate geometry aids in visualizing and solving problems relating the circle to axes. For example, ensuring the circle does not touch the axes involves understanding the circle's constraints and positioning. It's essential to know how the radius and center relate to each given point or line within the coordinate plane to solve intersecting problems effectively.