GSEB Class 11 Maths Solutions Chapter 11 Conic Sections Exercise 11.1

In Each of the Following Questions 1 to 5, Find the Equation of the Circle With

Question 1. Centre (0, 2) and radius 2.
Answer: The circle's center is (0, 2), and its radius is 2. The formula for the circle's equation is \( (x-h)^2 + (y-k)^2 = r^2 \). Using the given values, we get \( (x - 0)^2 + (y - 2)^2 = 2^2 \). This simplifies to \( x^2 + y^2 - 4y + 4 = 4 \), which further simplifies to \( x^2 + y^2 - 4y = 0 \).
In simple words: The center of the circle is (0, 2) and its radius is 2. We use a special formula to write the circle's equation. After putting in the numbers and simplifying, the equation becomes \( x^2 + y^2 - 4y = 0 \).

Exam Tip: Remember the standard form of a circle's equation: \( (x-h)^2 + (y-k)^2 = r^2 \), where (h,k) is the center and r is the radius.

Question 2. Centre (- 2, 3) and radius 4.
Answer: The circle's center is (- 2, 3), and its radius is 4. The formula for the circle's equation is \( (x-h)^2 + (y-k)^2 = r^2 \). By substituting the provided values, we get \( (x + 2)^2 + (y - 3)^2 = 4^2 \). Expanding this equation, we obtain \( x^2 + 4x + 4 + y^2 - 6y + 9 = 16 \). Rearranging the terms, this results in \( x^2 + y^2 + 4x - 6y + 13 - 16 = 0 \), which simplifies to \( x^2 + y^2 + 4x - 6y - 3 = 0 \).
In simple words: The circle's center is (- 2, 3) and its radius is 4. We use the circle equation formula. After putting in the numbers and expanding, the equation becomes \( x^2 + y^2 + 4x - 6y - 3 = 0 \).

Exam Tip: Pay careful attention to signs when substituting negative coordinates into the standard equation, especially for \( (x-h)^2 \) and \( (y-k)^2 \).

Question 3. Centre \( \left(\frac{1}{2}, \frac{1}{4}\right) \) and radius \( = \frac{1}{12} \).
Answer: The center of this circle is \( \left(\frac{1}{2}, \frac{1}{4}\right) \) and its radius is \( \frac{1}{12} \). The general equation for a circle is \( (x-h)^2 + (y-k)^2 = r^2 \). Substituting these values gives us \( \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \left(\frac{1}{12}\right)^2 \). This simplifies to \( \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{144} \). To remove fractions, we multiply the entire equation by 144, resulting in \( 144 \left(x^2 - x + \frac{1}{4}\right) + 144 \left(y^2 - \frac{1}{2}y + \frac{1}{16}\right) = 1 \). This further expands to \( 144x^2 - 144x + 36 + 144y^2 - 72y + 9 = 1 \). Combining like terms, we get \( 144x^2 + 144y^2 - 144x - 72y + 45 = 1 \). Finally, moving the constant to the left side and simplifying by dividing by 4, we arrive at \( 144x^2 + 144y^2 - 144x - 72y + 44 = 0 \), which becomes \( 36x^2 + 36y^2 - 36x - 18y + 11 = 0 \).
In simple words: The circle's center is \( \left(\frac{1}{2}, \frac{1}{4}\right) \) and its radius is \( \frac{1}{12} \). We use the circle's equation formula. After putting in the numbers, expanding, and simplifying to remove fractions, the final equation becomes \( 36x^2 + 36y^2 - 36x - 18y + 11 = 0 \).

Exam Tip: When dealing with fractional coordinates or radii, remember to clear denominators by multiplying the entire equation by the least common multiple of the denominators to simplify.

Question 4. Centre (1, 1) and radius \( \sqrt{2} \).
Answer: The center of this circle is (1, 1), and its radius is \( \sqrt{2} \). The equation for a circle is \( (x-h)^2 + (y-k)^2 = r^2 \). Plugging in the given center and radius values, we get \( (x - 1)^2 + (y - 1)^2 = (\sqrt{2})^2 \). This simplifies to \( x^2 - 2x + 1 + y^2 - 2y + 1 = 2 \). Combining terms, we have \( x^2 + y^2 - 2x - 2y + 2 = 2 \). Finally, subtracting 2 from both sides yields \( x^2 + y^2 - 2x - 2y = 0 \).
In simple words: The circle has its center at (1, 1) and a radius of \( \sqrt{2} \). Using the general formula for a circle, we substitute the values. After expanding and simplifying, the equation becomes \( x^2 + y^2 - 2x - 2y = 0 \).

Exam Tip: Remember that \( (\sqrt{a})^2 = a \). Be careful with the expansion of squared binomials like \( (x-1)^2 \), which is \( x^2 - 2x + 1 \).

Question 5. Centre (- a, - b) and radius \( \sqrt{a^{2}-b^{2}} \).
Answer: The center of the circle is (- a, - b), and its radius is \( \sqrt{a^{2}-b^{2}} \). The equation for a circle is \( (x-h)^2 + (y-k)^2 = r^2 \). By substituting these values, we derive \( (x - (-a))^2 + (y - (-b))^2 = (\sqrt{a^2-b^2})^2 \). This simplifies to \( (x + a)^2 + (y + b)^2 = a^2 - b^2 \). Expanding the terms, we get \( x^2 + 2ax + a^2 + y^2 + 2by + b^2 = a^2 - b^2 \). Rearranging and simplifying leads to \( x^2 + y^2 + 2ax + 2by + a^2 + b^2 - a^2 + b^2 = 0 \). This final simplification gives us \( x^2 + y^2 + 2ax + 2by + 2b^2 = 0 \).
In simple words: The circle's center is (- a, - b) with a radius of \( \sqrt{a^{2}-b^{2}} \). Applying the standard circle equation formula and expanding the terms, we get the equation \( x^2 + y^2 + 2ax + 2by + 2b^2 = 0 \).

Exam Tip: When the center coordinates are algebraic expressions (like -a, -b), ensure accurate substitution and careful expansion to avoid errors in the final equation.

Solutions to Questions 1 to 5:

In Each of the Following Questions 6 to 9, Find the Centre and Radius of the Circle:

Question 6. \( (x + 5)^2 + (y - 3)^2 = 36 \)
Answer: We need to find the center and radius of the circle given by the equation \( (x + 5)^2 + (y - 3)^2 = 36 \). We can compare this equation with the standard form of a circle's equation, which is \( (x - h)^2 + (y - k)^2 = r^2 \). From this comparison, we see that \( -h = 5 \), which means \( h = -5 \). Also, \( -k = -3 \), so \( k = 3 \). Finally, \( r^2 = 36 \), which implies that the radius \( r = \sqrt{36} = 6 \). Therefore, the circle's center is (- 5, 3), and its radius is 6.
In simple words: To find the center and radius, we match the given equation \( (x + 5)^2 + (y - 3)^2 = 36 \) to the standard circle formula \( (x - h)^2 + (y - k)^2 = r^2 \). This shows the center is (-5, 3) and the radius is 6.

Exam Tip: Carefully identify h, k, and r by matching signs in the standard equation \( (x-h)^2 + (y-k)^2 = r^2 \). For \( (x+5)^2 \), think of it as \( (x-(-5))^2 \), so \( h = -5 \).

Question 7. \( x^2 + y^2 - 4x - 8y - 45 = 0 \)
Answer: The given equation of the circle is \( x^2 + y^2 - 4x - 8y = 45 \). To find the center and radius, we will complete the square for both the \( x \) and \( y \) terms. We rearrange the terms as \( (x^2 - 4x) + (y^2 - 8y) = 45 \). For the \( x \) terms, we add \( \left(\frac{-4}{2}\right)^2 = 4 \) to complete the square, and for the \( y \) terms, we add \( \left(\frac{-8}{2}\right)^2 = 16 \). We must also add these values to the right side of the equation to maintain balance: \( (x^2 - 4x + 4) + (y^2 - 8y + 16) = 45 + 4 + 16 \). This transforms the equation into \( (x - 2)^2 + (y - 4)^2 = 65 \). Comparing this with the standard circle equation \( (x - h)^2 + (y - k)^2 = r^2 \), we determine that the center is (2, 4) and the radius \( r = \sqrt{65} \).
In simple words: To find the center and radius for \( x^2 + y^2 - 4x - 8y = 45 \), we complete the square. By adding 4 for \( x \) and 16 for \( y \) to both sides, the equation becomes \( (x - 2)^2 + (y - 4)^2 = 65 \). Thus, the center is (2, 4) and the radius is \( \sqrt{65} \).

Exam Tip: Completing the square is a key technique for converting a general quadratic equation of a circle into its standard form, which easily reveals the center and radius.

Question 8. \( x^2 + y^2 - 8x + 10y - 12 = 0 \)
Answer: The circle's equation is \( x^2 + y^2 - 8x + 10y - 12 = 0 \). To find its center and radius, we complete the square for the \( x \) and \( y \) terms. First, regroup the terms: \( (x^2 - 8x) + (y^2 + 10y) = 12 \). For \( x \), add \( \left(\frac{-8}{2}\right)^2 = 16 \). For \( y \), add \( \left(\frac{10}{2}\right)^2 = 25 \). Add these values to both sides: \( (x^2 - 8x + 16) + (y^2 + 10y + 25) = 12 + 16 + 25 \). This yields \( (x - 4)^2 + (y + 5)^2 = 53 \). Comparing this to the standard form \( (x - h)^2 + (y - k)^2 = r^2 \), we identify the center as (4, - 5) and the radius as \( \sqrt{53} \).
In simple words: For the equation \( x^2 + y^2 - 8x + 10y - 12 = 0 \), we complete the square. By adding 16 for \( x \) and 25 for \( y \) to both sides, the equation becomes \( (x - 4)^2 + (y + 5)^2 = 53 \). Thus, the center is (4, -5) and the radius is \( \sqrt{53} \).

Exam Tip: Always double-check your arithmetic, especially when adding constants to both sides of the equation while completing the square, to avoid simple calculation errors.

Question 9. \( 2x^2 + 2y^2 - x = 0 \)
Answer: The given equation for the circle is \( 2x^2 + 2y^2 - x = 0 \). To transform this into the standard form \( (x - h)^2 + (y - k)^2 = r^2 \), we first divide the entire equation by 2 to make the coefficients of \( x^2 \) and \( y^2 \) equal to 1: \( x^2 + y^2 - \frac{x}{2} = 0 \). Now, we complete the square for the \( x \) terms. We rewrite the equation as \( \left(x^2 - \frac{x}{2}\right) + y^2 = 0 \). To complete the square for \( x \), we add \( \left(\frac{-1/2}{2}\right)^2 = \left(-\frac{1}{4}\right)^2 = \frac{1}{16} \) to both sides. So, we have \( \left(x^2 - \frac{x}{2} + \frac{1}{16}\right) + y^2 = \frac{1}{16} \). This can be written as \( \left(x - \frac{1}{4}\right)^2 + (y - 0)^2 = \left(\frac{1}{4}\right)^2 \). Comparing this with the standard equation, the center of the circle is \( \left(\frac{1}{4}, 0\right) \) and the radius is \( \frac{1}{4} \).
In simple words: For the equation \( 2x^2 + 2y^2 - x = 0 \), we first divide by 2 to get \( x^2 + y^2 - \frac{x}{2} = 0 \). Then, we complete the square for the \( x \) terms by adding \( \frac{1}{16} \) to both sides. This gives us \( \left(x - \frac{1}{4}\right)^2 + y^2 = \left(\frac{1}{4}\right)^2 \). Therefore, the center is \( \left(\frac{1}{4}, 0\right) \) and the radius is \( \frac{1}{4} \).

Exam Tip: Always ensure the coefficients of \( x^2 \) and \( y^2 \) are 1 before attempting to complete the square. If they are not, divide the entire equation by that coefficient first.

Solutions to Questions 6 to 9:

Question 10. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line \( 4x + y = 16 \).
Answer: Let the general equation of the circle be \( (x - h)^2 + (y - k)^2 = r^2 \) (1). The problem states that the circle passes through points (4, 1) and (6, 5). When we substitute (4, 1) into equation (1), we get \( (4 - h)^2 + (1 - k)^2 = r^2 \), which expands to \( 16 - 8h + h^2 + 1 - 2k + k^2 = r^2 \), or \( h^2 + k^2 - 8h - 2k + 17 = r^2 \) (2). Similarly, substituting (6, 5) into equation (1) gives \( (6 - h)^2 + (5 - k)^2 = r^2 \), which expands to \( 36 - 12h + h^2 + 25 - 10k + k^2 = r^2 \), or \( h^2 + k^2 - 12h - 10k + 61 = r^2 \) (3).

The center (h, k) also lies on the line \( 4x + y = 16 \). So, \( 4h + k = 16 \) (4).

Now, we subtract equation (3) from equation (2) to eliminate \( r^2 \):
\( (h^2 + k^2 - 8h - 2k + 17) - (h^2 + k^2 - 12h - 10k + 61) = r^2 - r^2 \)
\( -8h - 2k + 17 + 12h + 10k - 61 = 0 \)
\( 4h + 8k - 44 = 0 \)
Dividing by 4, we obtain \( h + 2k = 11 \) (5).

We now have a system of two linear equations for h and k:
(4) \( 4h + k = 16 \)
(5) \( h + 2k = 11 \)
Multiply equation (5) by 4: \( 4(h + 2k) = 4(11) \), which is \( 4h + 8k = 44 \).
Subtract equation (4) from this new equation: \( (4h + 8k) - (4h + k) = 44 - 16 \)
\( 7k = 28 \)
This implies \( k = 4 \).

Substitute \( k = 4 \) back into equation (5): \( h + 2(4) = 11 \)
\( h + 8 = 11 \)
\( h = 3 \).

Now that we have the center (h, k) = (3, 4), we can find \( r^2 \) by substituting these values into equation (2):
\( r^2 = (3)^2 + (4)^2 - 8(3) - 2(4) + 17 \)
\( r^2 = 9 + 16 - 24 - 8 + 17 \)
\( r^2 = 42 - 32 \)
\( r^2 = 10 \).

Therefore, the equation of the circle is \( (x - 3)^2 + (y - 4)^2 = 10 \). Expanding this, we get \( x^2 - 6x + 9 + y^2 - 8y + 16 = 10 \), which simplifies to \( x^2 + y^2 - 6x - 8y + 25 - 10 = 0 \), or \( x^2 + y^2 - 6x - 8y + 15 = 0 \).
In simple words: First, we write the general circle equation \( (x - h)^2 + (y - k)^2 = r^2 \). We then use the two given points, (4, 1) and (6, 5), to create two equations for \( h, k, \) and \( r^2 \). Since the center (h, k) is on the line \( 4x + y = 16 \), we also get \( 4h + k = 16 \). By solving these simultaneous equations, we find \( h = 3 \) and \( k = 4 \). Substituting these back into one of the circle equations, we calculate \( r^2 = 10 \). So, the final equation of the circle is \( (x - 3)^2 + (y - 4)^2 = 10 \), which can be expanded to \( x^2 + y^2 - 6x - 8y + 15 = 0 \).

Exam Tip: When a circle passes through multiple points and its center lies on a line, form a system of equations. Two equations from the points (equating \( r^2 \)) and one from the center-on-line condition will help solve for h, k, and r.

Question 11. Find the equation of the circle passing through the points (2, 3) and (- 1, 1) and whose centre is on the line \( x - 3y - 11 = 0 \).
Answer: Let the general equation of the circle be \( (x - h)^2 + (y - k)^2 = r^2 \) (1). The circle passes through points (2, 3) and (- 1, 1).

Substituting (2, 3) into (1): \( (2 - h)^2 + (3 - k)^2 = r^2 \), which expands to \( 4 - 4h + h^2 + 9 - 6k + k^2 = r^2 \), or \( h^2 + k^2 - 4h - 6k + 13 = r^2 \) (2).

Substituting (- 1, 1) into (1): \( (-1 - h)^2 + (1 - k)^2 = r^2 \), which expands to \( 1 + 2h + h^2 + 1 - 2k + k^2 = r^2 \), or \( h^2 + k^2 + 2h - 2k + 2 = r^2 \) (3).

The center (h, k) lies on the line \( x - 3y - 11 = 0 \). So, \( h - 3k - 11 = 0 \) (4).

Subtract equation (3) from equation (2) to eliminate \( r^2 \):
\( (h^2 + k^2 - 4h - 6k + 13) - (h^2 + k^2 + 2h - 2k + 2) = 0 \)
\( -4h - 6k + 13 - 2h + 2k - 2 = 0 \)
\( -6h - 4k + 11 = 0 \)
\( 6h + 4k - 11 = 0 \) (5).

Now we have a system of two linear equations for h and k:
(4) \( h - 3k = 11 \)
(5) \( 6h + 4k = 11 \)
From (4), \( h = 3k + 11 \). Substitute this into (5):
\( 6(3k + 11) + 4k = 11 \)
\( 18k + 66 + 4k = 11 \)
\( 22k = 11 - 66 \)
\( 22k = -55 \)
\( k = -\frac{55}{22} = -\frac{5}{2} \).

Now, substitute \( k = -\frac{5}{2} \) back into \( h = 3k + 11 \):
\( h = 3\left(-\frac{5}{2}\right) + 11 = -\frac{15}{2} + \frac{22}{2} = \frac{7}{2} \).

So, the center is \( \left(\frac{7}{2}, -\frac{5}{2}\right) \). Now, substitute h and k values into equation (2) to find \( r^2 \):
\( r^2 = \left(2 - \frac{7}{2}\right)^2 + \left(3 - \left(-\frac{5}{2}\right)\right)^2 \)
\( r^2 = \left(\frac{4 - 7}{2}\right)^2 + \left(\frac{6 + 5}{2}\right)^2 \)
\( r^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{11}{2}\right)^2 \)
\( r^2 = \frac{9}{4} + \frac{121}{4} = \frac{130}{4} = \frac{65}{2} \).

Therefore, the equation of the circle is \( \left(x - \frac{7}{2}\right)^2 + \left(y - \left(-\frac{5}{2}\right)\right)^2 = \frac{65}{2} \).
\( \left(x - \frac{7}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2 = \frac{65}{2} \).
Expanding this: \( x^2 - 7x + \frac{49}{4} + y^2 + 5y + \frac{25}{4} = \frac{65}{2} \).
\( x^2 + y^2 - 7x + 5y + \frac{49}{4} + \frac{25}{4} - \frac{65 \times 2}{4} = 0 \).
\( x^2 + y^2 - 7x + 5y + \frac{49 + 25 - 130}{4} = 0 \).
\( x^2 + y^2 - 7x + 5y + \frac{74 - 130}{4} = 0 \).
\( x^2 + y^2 - 7x + 5y - \frac{56}{4} = 0 \).
\( x^2 + y^2 - 7x + 5y - 14 = 0 \).
In simple words: First, write the general circle equation. Using the two given points, (2, 3) and (-1, 1), we form two equations for \( h, k, \) and \( r^2 \). We also use the line \( x - 3y - 11 = 0 \) to get a third equation involving \( h \) and \( k \). Solving these equations, we find the center coordinates \( h = \frac{7}{2} \) and \( k = -\frac{5}{2} \). Then, we calculate \( r^2 = \frac{65}{2} \). Finally, substitute these values back into the circle equation to get \( \left(x - \frac{7}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2 = \frac{65}{2} \), which simplifies to \( x^2 + y^2 - 7x + 5y - 14 = 0 \).

Exam Tip: Keep your algebraic calculations organized, especially when dealing with fractions. A small error in solving simultaneous equations will lead to an incorrect final equation for the circle.

Question 12. Find the equation of the circle with radius 5, whose centre lies on x-axis and passes through the point (2, 3).
Answer: Let the equation of the circle be \( (x - h)^2 + (y - k)^2 = r^2 \) (1). We are given that the radius \( r = 5 \), so \( r^2 = 25 \). The center lies on the x-axis, which means the y-coordinate of the center is 0; thus, \( k = 0 \).

Substituting \( r^2 = 25 \) and \( k = 0 \) into equation (1), it becomes \( (x - h)^2 + y^2 = 25 \).

The circle passes through the point (2, 3). So, we substitute \( x = 2 \) and \( y = 3 \) into the equation:
\( (2 - h)^2 + 3^2 = 25 \)
\( (2 - h)^2 + 9 = 25 \)
\( (2 - h)^2 = 25 - 9 \)
\( (2 - h)^2 = 16 \).
Taking the square root of both sides gives \( 2 - h = \pm 4 \).

This leads to two possible cases for \( h \):
Case 1: \( 2 - h = 4 \implies h = 2 - 4 \implies h = -2 \).
Case 2: \( 2 - h = -4 \implies h = 2 + 4 \implies h = 6 \).

Therefore, there are two possible circles:
If \( h = -2 \), the center is (-2, 0) and the equation is \( (x - (-2))^2 + y^2 = 25 \), which simplifies to \( (x + 2)^2 + y^2 = 25 \). Expanding this, we get \( x^2 + 4x + 4 + y^2 = 25 \), so \( x^2 + y^2 + 4x - 21 = 0 \).

If \( h = 6 \), the center is (6, 0) and the equation is \( (x - 6)^2 + y^2 = 25 \). Expanding this, we get \( x^2 - 12x + 36 + y^2 = 25 \), so \( x^2 + y^2 - 12x + 11 = 0 \).

Thus, the required equations of the circles are \( x^2 + y^2 + 4x - 21 = 0 \) and \( x^2 + y^2 - 12x + 11 = 0 \).
In simple words: We begin with the standard circle equation \( (x - h)^2 + (y - k)^2 = r^2 \). We know the radius \( r = 5 \), so \( r^2 = 25 \). Since the center is on the x-axis, \( k = 0 \). Substituting these into the equation and using the point (2, 3) the circle passes through, we find two possible values for \( h \): \( -2 \) and \( 6 \). This gives us two possible equations for the circle: \( x^2 + y^2 + 4x - 21 = 0 \) and \( x^2 + y^2 - 12x + 11 = 0 \).

Exam Tip: When the center lies on an axis, one of the coordinates (h or k) will be zero. Remember that taking a square root results in both positive and negative values, which can lead to multiple solutions for the circle's equation.

Question 13. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the co-ordinate axes.
Answer: Let the circle pass through the origin (0, 0) and make intercepts of length \( a \) on the x-axis and \( b \) on the y-axis. This means the circle passes through points A(a, 0) and B(0, b). The line segment AB is a chord of the circle. The center of the circle is the midpoint of AB. Therefore, the center C is \( \left(\frac{a+0}{2}, \frac{0+b}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right) \).

The radius of the circle, denoted by OC, is the distance from the center C to the origin (0, 0).
\( \text{Radius} = OC = \sqrt{\left(\frac{a}{2} - 0\right)^2 + \left(\frac{b}{2} - 0\right)^2} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \frac{\sqrt{a^2 + b^2}}{2} \).

Now, using the standard equation of a circle \( (x - h)^2 + (y - k)^2 = r^2 \), with center \( (h, k) = \left(\frac{a}{2}, \frac{b}{2}\right) \) and radius \( r = \frac{\sqrt{a^2 + b^2}}{2} \), we get:
\( \left(x - \frac{a}{2}\right)^2 + \left(y - \frac{b}{2}\right)^2 = \left(\frac{\sqrt{a^2 + b^2}}{2}\right)^2 \)
\( x^2 - ax + \frac{a^2}{4} + y^2 - by + \frac{b^2}{4} = \frac{a^2 + b^2}{4} \).
Multiplying by 4 to eliminate denominators:
\( 4x^2 - 4ax + a^2 + 4y^2 - 4by + b^2 = a^2 + b^2 \).
Rearranging and simplifying, we get:
\( 4x^2 + 4y^2 - 4ax - 4by + a^2 + b^2 - a^2 - b^2 = 0 \).
\( 4x^2 + 4y^2 - 4ax - 4by = 0 \).
Dividing by 4, the final equation of the circle is \( x^2 + y^2 - ax - by = 0 \).
In simple words: When a circle passes through the origin (0, 0) and makes intercepts \( a \) and \( b \) on the coordinate axes, its center is at \( \left(\frac{a}{2}, \frac{b}{2}\right) \). The radius is the distance from this center to the origin, which is \( \frac{\sqrt{a^2 + b^2}}{2} \). Using the standard circle equation with these values, we expand and simplify to get the final equation: \( x^2 + y^2 - ax - by = 0 \).

Exam Tip: For a circle passing through the origin and making intercepts on the axes, its center will always be at \( \left(\frac{\text{x-intercept}}{2}, \frac{\text{y-intercept}}{2}\right) \). The origin is also a point on the circle.

Question 14. Find the equation of a circle with centre (2, 2) and which passes through the point (4, 5).
Answer: The center of the circle is given as C(2, 2). The circle passes through point P(4, 5). The radius of the circle is the distance between its center C and the point P on its circumference. We calculate the radius using the distance formula:
\( \text{Radius} = CP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( CP = \sqrt{(4 - 2)^2 + (5 - 2)^2} \)
\( CP = \sqrt{(2)^2 + (3)^2} \)
\( CP = \sqrt{4 + 9} \)
\( CP = \sqrt{13} \).
Thus, the radius \( r = \sqrt{13} \), and \( r^2 = 13 \).
Using the standard equation of a circle \( (x - h)^2 + (y - k)^2 = r^2 \), with center (h, k) = (2, 2) and \( r^2 = 13 \), the equation of the circle is \( (x - 2)^2 + (y - 2)^2 = 13 \).
In simple words: The center of the circle is (2, 2) and it passes through point (4, 5). We first find the radius by calculating the distance between these two points using the distance formula, which gives \( \sqrt{13} \). Then, we use the standard circle equation \( (x - h)^2 + (y - k)^2 = r^2 \) with the center (2, 2) and \( r^2 = 13 \) to get the equation \( (x - 2)^2 + (y - 2)^2 = 13 \).

Exam Tip: When given the center and a point on the circle, the distance formula is your direct path to finding the radius, which is essential for writing the circle's equation.

Question 15. Does the point (- 2.5, 3.5) lie inside, outside or on the circle \( x^2 + y^2 = 25 \)?
Answer: The equation of the circle is \( x^2 + y^2 = 25 \). From this equation, we can determine that the center of the circle C is (0, 0) and the radius \( r = \sqrt{25} = 5 \).

We need to find the position of the point P(- 2.5, 3.5) relative to this circle. To do this, we calculate the distance between the center C(0, 0) and the point P(- 2.5, 3.5). Using the distance formula:
\( CP = \sqrt{(-2.5 - 0)^2 + (3.5 - 0)^2} \)
\( CP = \sqrt{(-2.5)^2 + (3.5)^2} \)
\( CP = \sqrt{6.25 + 12.25} \)
\( CP = \sqrt{18.5} \).
Now, we compare this distance (CP) with the circle's radius (r).
\( CP = \sqrt{18.5} \approx 4.301 \).
Since \( \sqrt{18.5} < 5 \) (because \( 18.5 < 25 \)), it means that the distance from the center to the point P is less than the radius. Therefore, the point P lies inside the circle.
In simple words: The circle \( x^2 + y^2 = 25 \) has its center at (0, 0) and a radius of 5. To check where point P(-2.5, 3.5) is, we find its distance from the center. This distance is \( \sqrt{18.5} \). Since \( \sqrt{18.5} \) is smaller than the radius (5), the point P lies inside the circle.

Exam Tip: To determine if a point is inside, outside, or on a circle, compare its distance from the center (d) to the radius (r). If d < r, it's inside; if d = r, it's on; if d > r, it's outside.