Samacheer Kalvi Class 11 Business Maths Solutions Chapter 3 Analytical Geometry Exercise 3.4

Question 1. Find the equation of the following circles having
(i) the centre (3, 5) and radius 5 units.
(ii) the centre (0, 0) and radius 2 units.
Answer:
(i) The general equation for a circle with centre \( (h, k) \) and radius \( r \) is given by \( (x - h)^2 + (y - k)^2 = r^2 \).
Here, the centre \( (h, k) \) is \( (3, 5) \) and the radius \( r \) is 5.
So, we substitute these values into the equation:
\( (x - 3)^2 + (y - 5)^2 = 5^2 \)
Now, we expand the terms:
\( (x^2 - 6x + 9) + (y^2 - 10y + 25) = 25 \)
To simplify, we move 25 from the right side to the left:
\( x^2 - 6x + 9 + y^2 - 10y + 25 - 25 = 0 \)
\( \implies x^2 + y^2 - 6x - 10y + 9 = 0 \)

(ii) For a circle with its centre at the origin \( (0, 0) \) and radius \( r \), the equation is \( x^2 + y^2 = r^2 \).
Here, the centre is \( (0, 0) \) and the radius \( r \) is 2.
Substitute the radius into the equation:
\( x^2 + y^2 = 2^2 \)
\( x^2 + y^2 = 4 \)
\( \implies x^2 + y^2 - 4 = 0 \)
In simple words: To find a circle's equation, we use a special formula. For a circle not at the center, we put in where its center is and how big its radius is. For a circle starting at the very middle (origin), we just need to know its radius.

🎯 Exam Tip: Remember the two main forms of a circle's equation: \( (x - h)^2 + (y - k)^2 = r^2 \) for any centre \( (h,k) \) and \( x^2 + y^2 = r^2 \) specifically for a centre at the origin \( (0,0) \).

Question 2. Find the centre and radius of the circle
(i) \( x^2 + y^2 = 16 \)
(ii) \( x^2 + y^2 - 22x - 4y + 25 = 0 \)
(iii) \( 5x^2 + 5y^2 + 4x - 8y - 16 = 0 \)
(iv) \( (x + 2) (x - 5) + (y - 2) (y - 1) = 0 \)
Answer:
(i) The given equation is \( x^2 + y^2 = 16 \).
This equation is in the form \( x^2 + y^2 = r^2 \), which represents a circle with its centre at the origin \( (0, 0) \).
By comparing, we can see that \( r^2 = 16 \).
Taking the square root, we get \( r = \sqrt{16} = 4 \).
Therefore, the centre of this circle is \( (0, 0) \) and its radius is 4 units.

(ii) The given equation is \( x^2 + y^2 - 22x - 4y + 25 = 0 \).
We compare this with the general equation of a circle: \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
From comparison, we get:
\( 2g = -22 \implies g = -11 \)
\( 2f = -4 \implies f = -2 \)
\( c = 25 \)
The centre of the circle is \( (-g, -f) \). So, the centre is \( (-(-11), -(-2)) = (11, 2) \).
The radius \( r \) is calculated using the formula \( r = \sqrt{g^2 + f^2 - c} \).
\( r = \sqrt{(-11)^2 + (-2)^2 - 25} \)
\( r = \sqrt{121 + 4 - 25} \)
\( r = \sqrt{100} \)
\( r = 10 \)
Therefore, the centre is \( (11, 2) \) and the radius is 10 units.

(iii) The given equation is \( 5x^2 + 5y^2 + 4x - 8y - 16 = 0 \).
For a circle equation, the coefficients of \( x^2 \) and \( y^2 \) must be 1. So, we divide the entire equation by 5:
\( \frac{5x^2}{5} + \frac{5y^2}{5} + \frac{4x}{5} - \frac{8y}{5} - \frac{16}{5} = 0 \)
\( \implies x^2 + y^2 + \frac{4}{5}x - \frac{8}{5}y - \frac{16}{5} = 0 \)
Now, we compare this with the general equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
From comparison, we get:
\( 2g = \frac{4}{5} \implies g = \frac{4}{5 \times 2} = \frac{4}{10} = \frac{2}{5} \)
\( 2f = -\frac{8}{5} \implies f = -\frac{8}{5 \times 2} = -\frac{8}{10} = -\frac{4}{5} \)
\( c = -\frac{16}{5} \)
The centre of the circle is \( (-g, -f) \). So, the centre is \( \left(-\frac{2}{5}, -(-\frac{4}{5})\right) = \left(-\frac{2}{5}, \frac{4}{5}\right) \).
The radius \( r \) is calculated using the formula \( r = \sqrt{g^2 + f^2 - c} \).
\( r = \sqrt{\left(\frac{2}{5}\right)^2 + \left(-\frac{4}{5}\right)^2 - \left(-\frac{16}{5}\right)} \)
\( r = \sqrt{\frac{4}{25} + \frac{16}{25} + \frac{16}{5}} \)
To add the fractions, find a common denominator, which is 25:
\( r = \sqrt{\frac{4}{25} + \frac{16}{25} + \frac{16 \times 5}{25}} \)
\( r = \sqrt{\frac{4 + 16 + 80}{25}} \)
\( r = \sqrt{\frac{100}{25}} \)
\( r = \sqrt{4} \)
\( r = 2 \)
Therefore, the centre is \( \left(-\frac{2}{5}, \frac{4}{5}\right) \) and the radius is 2 units.

(iv) The given equation is \( (x + 2) (x - 5) + (y - 2) (y - 1) = 0 \).
First, we need to expand the products:
\( x^2 - 5x + 2x - 10 + y^2 - y - 2y + 2 = 0 \)
\( \implies x^2 + y^2 - 3x - 3y - 8 = 0 \)
Now, we compare this with the general equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
From comparison, we get:
\( 2g = -3 \implies g = -\frac{3}{2} \)
\( 2f = -3 \implies f = -\frac{3}{2} \)
\( c = -8 \)
The centre of the circle is \( (-g, -f) \). So, the centre is \( \left(-\left(-\frac{3}{2}\right), -\left(-\frac{3}{2}\right)\right) = \left(\frac{3}{2}, \frac{3}{2}\right) \).
The radius \( r \) is calculated using the formula \( r = \sqrt{g^2 + f^2 - c} \).
\( r = \sqrt{\left(-\frac{3}{2}\right)^2 + \left(-\frac{3}{2}\right)^2 - (-8)} \)
\( r = \sqrt{\frac{9}{4} + \frac{9}{4} + 8} \)
\( r = \sqrt{\frac{18}{4} + 8} \)
\( r = \sqrt{\frac{9}{2} + \frac{16}{2}} \)
\( r = \sqrt{\frac{25}{2}} \)
\( r = \frac{5}{\sqrt{2}} \)
Therefore, the centre is \( \left(\frac{3}{2}, \frac{3}{2}\right) \) and the radius is \( \frac{5}{\sqrt{2}} \) units.
In simple words: To find the centre and radius from a circle's equation, we match it with a general form. If the equation is simple like \( x^2 + y^2 = r^2 \), the centre is at \( (0,0) \). If it has \( x \) and \( y \) terms, we use special formulas involving \( g, f, \) and \( c \) to find the centre and radius. Sometimes, we need to make the \( x^2 \) and \( y^2 \) terms equal to 1 first.

🎯 Exam Tip: Always remember to check if the coefficients of \( x^2 \) and \( y^2 \) are 1 before comparing with the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \). If not, divide the entire equation by that coefficient.

Question 3. Find the equation of the circle whose centre is (-3, -2) and having circumference 16π.
Answer: The centre of the circle is given as \( (h, k) = (-3, -2) \).
The circumference of the circle is given as \( 16\pi \).
We know the formula for the circumference of a circle is \( C = 2\pi r \).
So, we have:
\( 2\pi r = 16\pi \)
To find the radius \( r \), we can divide both sides by \( 2\pi \):
\( r = \frac{16\pi}{2\pi} \)
\( \implies r = 8 \)
Now we have the centre \( (h, k) = (-3, -2) \) and the radius \( r = 8 \).
The equation of a circle with centre \( (h, k) \) and radius \( r \) is \( (x - h)^2 + (y - k)^2 = r^2 \).
Substitute the values:
\( (x - (-3))^2 + (y - (-2))^2 = 8^2 \)
\( (x + 3)^2 + (y + 2)^2 = 64 \)
Now, we expand the terms:
\( (x^2 + 6x + 9) + (y^2 + 4y + 4) = 64 \)
To find the general form, move 64 to the left side:
\( x^2 + 6x + 9 + y^2 + 4y + 4 - 64 = 0 \)
\( \implies x^2 + y^2 + 6x + 4y + 13 - 64 = 0 \)
\( \implies x^2 + y^2 + 6x + 4y - 51 = 0 \)
This is the required equation of the circle.
In simple words: First, we use the given circumference to find how big the circle's radius is. Then, knowing the centre and this radius, we put these numbers into the standard circle equation formula to get the final equation.

🎯 Exam Tip: Always ensure you use the correct formula for circumference or area to derive the radius, as this is a crucial first step for many circle problems.

Question 4. Find the equation of the circle whose centre is (2, 3) and which passes through (1, 4).
Answer: The centre of the circle is given as \( (h, k) = (2, 3) \).
The circle passes through the point \( (x_1, y_1) = (1, 4) \).
The radius \( r \) of the circle is the distance between the centre \( (2, 3) \) and the point \( (1, 4) \) through which it passes.
We use the distance formula: \( r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
Let \( (x_1, y_1) = (2, 3) \) and \( (x_2, y_2) = (1, 4) \).
\( r = \sqrt{(1 - 2)^2 + (4 - 3)^2} \)
\( r = \sqrt{(-1)^2 + (1)^2} \)
\( r = \sqrt{1 + 1} \)
\( r = \sqrt{2} \)
Now we have the centre \( (h, k) = (2, 3) \) and the radius \( r = \sqrt{2} \).
The equation of a circle with centre \( (h, k) \) and radius \( r \) is \( (x - h)^2 + (y - k)^2 = r^2 \).
Substitute the values:
\( (x - 2)^2 + (y - 3)^2 = (\sqrt{2})^2 \)
\( (x - 2)^2 + (y - 3)^2 = 2 \)
Now, we expand the terms:
\( (x^2 - 4x + 4) + (y^2 - 6y + 9) = 2 \)
To find the general form, move 2 to the left side:
\( x^2 - 4x + 4 + y^2 - 6y + 9 - 2 = 0 \)
\( \implies x^2 + y^2 - 4x - 6y + 13 - 2 = 0 \)
\( \implies x^2 + y^2 - 4x - 6y + 11 = 0 \)
This is the required equation of the circle. The distance formula is essential for finding the radius in such cases.
In simple words: We are given the middle point (centre) of the circle and one point it touches. We use the distance formula to find how long the radius is from the center to that point. Once we have the centre and radius, we put them into the basic circle equation formula.

🎯 Exam Tip: When a circle passes through a point, the distance from the centre to that point is always the radius. This is a common way to find the radius when it's not directly given.

Question 5. Find the equation of the circle passing through the points (0, 1), (4, 3) and (1, -1).
Answer: Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \) (Equation 1).
Since the circle passes through the point \( (0, 1) \), we substitute these coordinates into Equation 1:
\( (0)^2 + (1)^2 + 2g(0) + 2f(1) + c = 0 \)
\( 1 + 2f + c = 0 \)
\( \implies 2f + c = -1 \) (Equation 2)

Next, the circle passes through the point \( (4, 3) \), so we substitute these coordinates into Equation 1:
\( (4)^2 + (3)^2 + 2g(4) + 2f(3) + c = 0 \)
\( 16 + 9 + 8g + 6f + c = 0 \)
\( 25 + 8g + 6f + c = 0 \)
\( \implies 8g + 6f + c = -25 \) (Equation 3)

Finally, the circle passes through the point \( (1, -1) \), so we substitute these coordinates into Equation 1:
\( (1)^2 + (-1)^2 + 2g(1) + 2f(-1) + c = 0 \)
\( 1 + 1 + 2g - 2f + c = 0 \)
\( 2 + 2g - 2f + c = 0 \)
\( \implies 2g - 2f + c = -2 \) (Equation 4)

Now we have a system of three linear equations (2, 3, and 4) with three unknowns (g, f, c).

Subtract Equation 2 from Equation 4:
\( (2g - 2f + c) - (2f + c) = -2 - (-1) \)
\( 2g - 4f = -1 \) (Equation 5)

Subtract Equation 3 from Equation 4 (this is incorrect in the source, it subtracts 4 from 3 for some terms but then uses 4 for subtraction from 2 for 8g - 8f + 4c = -8. Let's follow the standard elimination process):

Multiply Equation 4 by 4 to make the 'g' coefficient 8:
\( 4 \times (2g - 2f + c) = 4 \times (-2) \)
\( 8g - 8f + 4c = -8 \) (Equation 6)

Now, subtract Equation 6 from Equation 3:
\( (8g + 6f + c) - (8g - 8f + 4c) = -25 - (-8) \)
\( 8g + 6f + c - 8g + 8f - 4c = -25 + 8 \)
\( 14f - 3c = -17 \) (Equation 7)

From Equation 2, we have \( c = -1 - 2f \). Substitute this into Equation 7:
\( 14f - 3(-1 - 2f) = -17 \)
\( 14f + 3 + 6f = -17 \)
\( 20f + 3 = -17 \)
\( 20f = -17 - 3 \)
\( 20f = -20 \)
\( f = -1 \)

Now substitute \( f = -1 \) back into Equation 2 to find \( c \):
\( 2(-1) + c = -1 \)
\( -2 + c = -1 \)
\( c = -1 + 2 \)
\( c = 1 \)

Now substitute \( f = -1 \) and \( c = 1 \) back into Equation 3 to find \( g \):
\( 8g + 6(-1) + 1 = -25 \)
\( 8g - 6 + 1 = -25 \)
\( 8g - 5 = -25 \)
\( 8g = -25 + 5 \)
\( 8g = -20 \)
\( g = -\frac{20}{8} \)
\( g = -\frac{5}{2} \)

Finally, substitute the values of \( g = -\frac{5}{2} \), \( f = -1 \), and \( c = 1 \) into the general equation of the circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( x^2 + y^2 + 2\left(-\frac{5}{2}\right)x + 2(-1)y + 1 = 0 \)
\( x^2 + y^2 - 5x - 2y + 1 = 0 \)
This is the required equation of the circle. This method involves solving a system of linear equations to find the circle's parameters.
In simple words: When a circle goes through three points, we put each point's coordinates into the circle's general equation. This gives us three math puzzles to solve together. By solving these, we find the special numbers (g, f, c) that define the circle. Once we have these numbers, we put them back into the general equation to get the circle's exact formula.

🎯 Exam Tip: This type of problem often leads to a system of linear equations. Be careful with calculations and algebraic manipulation to avoid errors when solving for g, f, and c.

Question 6. Find the equation of the circle on the line joining the points (1, 0), (0, 1), and having its centre on the line x + y = 1.
Answer: Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \) (Equation 1).
The circle passes through the point \( (1, 0) \). Substitute these coordinates into Equation 1:
\( (1)^2 + (0)^2 + 2g(1) + 2f(0) + c = 0 \)
\( 1 + 2g + c = 0 \)
\( \implies 2g + c = -1 \) (Equation 2)

The circle also passes through the point \( (0, 1) \). Substitute these coordinates into Equation 1:
\( (0)^2 + (1)^2 + 2g(0) + 2f(1) + c = 0 \)
\( 1 + 2f + c = 0 \)
\( \implies 2f + c = -1 \) (Equation 3)

The centre of the circle is \( (-g, -f) \). We are given that the centre lies on the line \( x + y = 1 \).
So, substitute the coordinates of the centre into the line equation:
\( (-g) + (-f) = 1 \)
\( \implies -g - f = 1 \) (Equation 4)

Now we have a system of four equations (though we only need three independent ones). Let's use Equations 2, 3, and 4.

Subtract Equation 3 from Equation 2:
\( (2g + c) - (2f + c) = -1 - (-1) \)
\( 2g - 2f = 0 \)
\( \implies g = f \) (Equation 5)

Now substitute \( g = f \) into Equation 4:
\( -f - f = 1 \)
\( -2f = 1 \)
\( f = -\frac{1}{2} \)

Since \( g = f \), we also have \( g = -\frac{1}{2} \).

Now substitute \( f = -\frac{1}{2} \) into Equation 3 to find \( c \):
\( 2\left(-\frac{1}{2}\right) + c = -1 \)
\( -1 + c = -1 \)
\( c = 0 \)

Finally, substitute the values of \( g = -\frac{1}{2} \), \( f = -\frac{1}{2} \), and \( c = 0 \) into the general equation of the circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( x^2 + y^2 + 2\left(-\frac{1}{2}\right)x + 2\left(-\frac{1}{2}\right)y + 0 = 0 \)
\( x^2 + y^2 - x - y = 0 \)
This is the required equation of the circle. Finding the relationship between g and f simplified the solution.
In simple words: We start with the general circle equation. Since the circle goes through two points, we use those points to make two equations. The extra hint is that the circle's middle point (centre) sits on a specific line, which gives us a third equation. We solve these equations to find the special numbers (g, f, c) and then build the final circle equation.

🎯 Exam Tip: When given conditions about the circle's centre (e.g., it lies on a line), always substitute the centre's coordinates \( (-g, -f) \) into the given line equation to form an additional relationship between g and f.

Question 7. If the lines x + y = 6 and x + 2y = 4 are diameters of the circle, and the circle passes through the point (2, 6) then find its equation.
Answer: The intersection point of any two diameters of a circle is its centre. So, we need to solve the given linear equations to find the centre \( (h, k) \).
Equation 1: \( x + y = 6 \)
Equation 2: \( x + 2y = 4 \)

Subtract Equation 1 from Equation 2:
\( (x + 2y) - (x + y) = 4 - 6 \)
\( y = -2 \)

Now substitute \( y = -2 \) into Equation 1:
\( x + (-2) = 6 \)
\( x - 2 = 6 \)
\( x = 8 \)
So, the centre of the circle is \( (h, k) = (8, -2) \).

The circle passes through the point \( (x_1, y_1) = (2, 6) \).
The radius \( r \) is the distance between the centre \( (8, -2) \) and the point \( (2, 6) \).
Using the distance formula: \( r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
Let \( (x_1, y_1) = (8, -2) \) and \( (x_2, y_2) = (2, 6) \).
\( r = \sqrt{(2 - 8)^2 + (6 - (-2))^2} \)
\( r = \sqrt{(-6)^2 + (6 + 2)^2} \)
\( r = \sqrt{(-6)^2 + (8)^2} \)
\( r = \sqrt{36 + 64} \)
\( r = \sqrt{100} \)
\( r = 10 \)
Now we have the centre \( (h, k) = (8, -2) \) and the radius \( r = 10 \).
The equation of a circle with centre \( (h, k) \) and radius \( r \) is \( (x - h)^2 + (y - k)^2 = r^2 \).
Substitute the values:
\( (x - 8)^2 + (y - (-2))^2 = 10^2 \)
\( (x - 8)^2 + (y + 2)^2 = 100 \)
Now, expand the terms:
\( (x^2 - 16x + 64) + (y^2 + 4y + 4) = 100 \)
To find the general form, move 100 to the left side:
\( x^2 - 16x + 64 + y^2 + 4y + 4 - 100 = 0 \)
\( \implies x^2 + y^2 - 16x + 4y + 68 - 100 = 0 \)
\( \implies x^2 + y^2 - 16x + 4y - 32 = 0 \)
This is the required equation of the circle. This problem highlights how the intersection of diameters gives the centre.
In simple words: First, we find the circle's middle point (centre) by figuring out where the two diameter lines cross. Then, we measure the distance from this centre to the point the circle passes through; this distance is the radius. Finally, we use the centre and radius in the standard circle equation formula to write down the circle's full equation.

🎯 Exam Tip: Remember that all diameters of a circle pass through its centre. Solving the equations of two diameters will always give you the coordinates of the circle's centre.

Question 8. Find the equation of the circle having (4, 7) and (-2, 5) as the extremities of a diameter.
Answer: Let the two given points be \( (x_1, y_1) = (4, 7) \) and \( (x_2, y_2) = (-2, 5) \).
These points are the ends of a diameter of the circle.
The equation of a circle when the endpoints of a diameter are \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \)
Substitute the given coordinates into this formula:
\( (x - 4)(x - (-2)) + (y - 7)(y - 5) = 0 \)
\( (x - 4)(x + 2) + (y - 7)(y - 5) = 0 \)
Now, expand both products:
\( (x^2 + 2x - 4x - 8) + (y^2 - 5y - 7y + 35) = 0 \)
\( (x^2 - 2x - 8) + (y^2 - 12y + 35) = 0 \)
Combine like terms to get the general form of the circle equation:
\( x^2 + y^2 - 2x - 12y - 8 + 35 = 0 \)
\( \implies x^2 + y^2 - 2x - 12y + 27 = 0 \)
This is the required equation of the circle. This formula provides a shortcut when diameter endpoints are known.
In simple words: If you know the two ends of a circle's diameter, there's a quick math trick to find its equation. You simply multiply \( (x - \text{first x-point}) \) by \( (x - \text{second x-point}) \), and add it to \( (y - \text{first y-point}) \) multiplied by \( (y - \text{second y-point}) \), then set the whole thing equal to zero.

🎯 Exam Tip: Memorizing the diameter form of the circle equation, \( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \), can save significant time compared to finding the centre (midpoint) and radius (half the distance) separately.

Question 9. Find the Cartesian equation of the circle whose parametric equations are x = 3 cos θ, y = 3 sin θ, 0 ≤ θ ≤ 2π.
Answer: We are given the parametric equations of the circle:
\( x = 3 \cos \theta \) (Equation 1)
\( y = 3 \sin \theta \) (Equation 2)
To find the Cartesian equation, we need to eliminate the parameter \( \theta \).
From Equation 1, we can write \( \cos \theta = \frac{x}{3} \).
From Equation 2, we can write \( \sin \theta = \frac{y}{3} \).
We know the fundamental trigonometric identity: \( \cos^2 \theta + \sin^2 \theta = 1 \).
Substitute the expressions for \( \cos \theta \) and \( \sin \theta \) into this identity:
\( \left(\frac{x}{3}\right)^2 + \left(\frac{y}{3}\right)^2 = 1 \)
\( \frac{x^2}{9} + \frac{y^2}{9} = 1 \)
Multiply the entire equation by 9 to clear the denominators:
\( 9 \left(\frac{x^2}{9}\right) + 9 \left(\frac{y^2}{9}\right) = 9 \times 1 \)
\( \implies x^2 + y^2 = 9 \)
This is the Cartesian equation of the circle. This method uses a key trigonometric identity to remove the parameter.
In simple words: When a circle is described using an angle (theta), we use a special math rule that says \( (\cos \theta)^2 + (\sin \theta)^2 \) always equals 1. By changing the x and y equations to fit this rule, we can get rid of the angle and find the circle's usual equation.

🎯 Exam Tip: For parametric equations involving sine and cosine, the key to finding the Cartesian equation is almost always to use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) after isolating \( \sin \theta \) and \( \cos \theta \).