Samacheer Kalvi Class 12 Maths Solutions Chapter 5 Two Dimensional Analytical Geometry II Exercise 5.1

Question 1. Obtain the equation of the circles with a radius of 5 cm and touching the x-axis at the origin in a general form.
Answer:The problem asks for the equation of circles with a radius of 5 cm that touch the x-axis at the origin. Since the circle touches the x-axis at the origin, its center will be at \( (0, \pm 5) \) and the radius will be 5 units. We can have two such circles, one with the center \( (0, 5) \) and another with \( (0, -5) \). The general equation of a circle with center \( (h, k) \) and radius \( r \) is \( (x - h)^2 + (y - k)^2 = r^2 \). For the circle with center \( (0, 5) \) and radius \( r = 5 \): \( (x - 0)^2 + (y - 5)^2 = 5^2 \) \( x^2 + y^2 - 10y + 25 = 25 \) \( x^2 + y^2 - 10y = 0 \) For the circle with center \( (0, -5) \) and radius \( r = 5 \): \( (x - 0)^2 + (y - (-5))^2 = 5^2 \) \( x^2 + (y + 5)^2 = 5^2 \) \( x^2 + y^2 + 10y + 25 = 25 \) \( x^2 + y^2 + 10y = 0 \) Combining both cases, the equation of the circles is \( x^2 + y^2 \pm 10y = 0 \).In simple words: If a circle touches the x-axis at the middle point (origin) and has a radius of 5, its center must be directly above or below the origin at \( (0, 5) \) or \( (0, -5) \). Using the circle formula, we get two possible equations: \( x^2 + y^2 - 10y = 0 \) for the top circle, and \( x^2 + y^2 + 10y = 0 \) for the bottom circle.

🎯 Exam Tip: Remember that "touching the x-axis at the origin" means the y-coordinate of the center will be equal to the radius (or its negative), and the x-coordinate of the center will be 0.

Question 2. Find the equation of the circle with centre (2, -1) and passing through the point (3, 6) in standard form.
Answer:We are given the center of the circle \( (h, k) = (2, -1) \). The circle passes through the point \( (x, y) = (3, 6) \). The standard form of the equation of a circle is \( (x - h)^2 + (y - k)^2 = r^2 \). First, we need to find the radius \( r \) of the circle. The distance from the center to the point it passes through is the radius. Substitute the center \( (2, -1) \) and the point \( (3, 6) \) into the distance formula to find \( r^2 \): \( r^2 = (3 - 2)^2 + (6 - (-1))^2 \) \( r^2 = (1)^2 + (6 + 1)^2 \) \( r^2 = 1^2 + 7^2 \) \( r^2 = 1 + 49 \)
\( \implies r^2 = 50 \) Now substitute the center \( (h, k) = (2, -1) \) and \( r^2 = 50 \) into the standard equation of the circle: \( (x - 2)^2 + (y - (-1))^2 = 50 \)
\( \implies (x - 2)^2 + (y + 1)^2 = 50 \) This is the equation of the circle in standard form.In simple words: We know the center of the circle and a point it goes through. We first calculate the squared distance between these two points to find \( r^2 \). Then, we use the standard circle equation \( (x - h)^2 + (y - k)^2 = r^2 \), plugging in the center and the \( r^2 \) value we just found.

🎯 Exam Tip: Remember that the radius is the distance from the center to any point on the circle. If you have the center and a point, use the distance formula to find the radius (or \( r^2 \)).

Question 3. Find the equation of circles that touch both the axes and pass-through (-4, -2).
Answer:If a circle touches both the x and y axes, its center will be \( (r, r) \) or \( (-r, r) \) or \( (r, -r) \) or \( (-r, -r) \), and its radius will be \( r \). Since the point \( (-4, -2) \) is in the third quadrant, the center of the circle must also be in the third quadrant. So, the center of the circle will be \( (-r, -r) \) and its radius will be \( r \). The equation of the circle is \( (x - (-r))^2 + (y - (-r))^2 = r^2 \), which simplifies to \( (x + r)^2 + (y + r)^2 = r^2 \). The circle passes through the point \( (-4, -2) \). Substitute these values into the equation: \( (-4 + r)^2 + (-2 + r)^2 = r^2 \) Expand the terms: \( (r - 4)^2 + (r - 2)^2 = r^2 \) \( (r^2 - 8r + 16) + (r^2 - 4r + 4) = r^2 \) Combine like terms: \( r^2 - 8r + 16 + r^2 - 4r + 4 - r^2 = 0 \)
\( \implies r^2 - 12r + 20 = 0 \) Factor the quadratic equation: \( (r - 2)(r - 10) = 0 \)
\( \implies r = 2 \) or \( r = 10 \) Case 1: When \( r = 2 \) The center is \( (-2, -2) \) and the radius is \( 2 \). The equation of the circle is \( (x + 2)^2 + (y + 2)^2 = 2^2 \) \( x^2 + 4x + 4 + y^2 + 4y + 4 = 4 \)
\( \implies x^2 + y^2 + 4x + 4y + 4 = 0 \) Case 2: When \( r = 10 \) The center is \( (-10, -10) \) and the radius is \( 10 \). The equation of the circle is \( (x + 10)^2 + (y + 10)^2 = 10^2 \) \( x^2 + 20x + 100 + y^2 + 20y + 100 = 100 \)
\( \implies x^2 + y^2 + 20x + 20y + 100 = 0 \) So there are two possible circles.In simple words: If a circle touches both the x and y axes, its center coordinates are the same as its radius (but with signs matching the quadrant). Since the circle goes through \( (-4, -2) \), its center must be \( (-r, -r) \). We put this into the circle equation and solve for \( r \), getting two possible values for the radius (2 and 10). Each radius gives a different circle equation.

🎯 Exam Tip: When a circle touches both axes and passes through a point in a specific quadrant, ensure the center's coordinates align with that quadrant's signs (e.g., \( (-r, -r) \) for the third quadrant).

Question 4. Find the equation of the circles with centre (2, 3) and passing through the intersection of the lines 3x - 2y - 1 = 0 and 4x + y - 27 = 0.
Answer:The center of the circle is given as \( (h, k) = (2, 3) \). The circle passes through the intersection point of the lines: Line 1: \( 3x - 2y - 1 = 0 \implies 3x - 2y = 1 \) (Equation 1) Line 2: \( 4x + y - 27 = 0 \implies 4x + y = 27 \) (Equation 2) To find the point of intersection, we can solve these two linear equations. Multiply Equation 2 by 2 to eliminate \( y \): \( 2 \times (4x + y) = 2 \times 27 \) \( 8x + 2y = 54 \) (Equation 3) Add Equation 1 and Equation 3: \( (3x - 2y) + (8x + 2y) = 1 + 54 \) \( 11x = 55 \)
\( \implies x = \frac{55}{11} = 5 \) Substitute \( x = 5 \) into Equation 1: \( 3(5) - 2y = 1 \) \( 15 - 2y = 1 \) \( -2y = 1 - 15 \) \( -2y = -14 \)
\( \implies y = \frac{-14}{-2} = 7 \) So, the point of intersection is \( (5, 7) \). This is the point the circle passes through. Now we have the center \( (h, k) = (2, 3) \) and a point on the circle \( (x, y) = (5, 7) \). We use the standard equation of a circle: \( (x - h)^2 + (y - k)^2 = r^2 \). First, find \( r^2 \) using the center and the point \( (5, 7) \): \( r^2 = (5 - 2)^2 + (7 - 3)^2 \) \( r^2 = (3)^2 + (4)^2 \) \( r^2 = 9 + 16 \)
\( \implies r^2 = 25 \) Now, substitute the center \( (2, 3) \) and \( r^2 = 25 \) into the standard equation: \( (x - 2)^2 + (y - 3)^2 = 25 \) Expand this equation to get the general form (if needed, but standard form is usually sufficient unless specified): \( (x^2 - 4x + 4) + (y^2 - 6y + 9) = 25 \) \( x^2 + y^2 - 4x - 6y + 13 = 25 \)
\( \implies x^2 + y^2 - 4x - 6y - 12 = 0 \) This is the equation of the circle.In simple words: We are given the center of the circle. To find its equation, we need the radius. The circle passes through where two lines cross. First, we solve the two line equations together to find that crossing point. Then, we use the distance formula between the circle's center and this crossing point to find the radius squared (\( r^2 \)). Finally, we plug the center and \( r^2 \) into the standard circle formula.

🎯 Exam Tip: When a circle passes through the intersection of lines, the first step is always to find that intersection point by solving the system of linear equations. This point then acts as a point on the circle for calculating the radius.

Question 5. Obtain the equation of the circle for which (3, 4) and (2, -7) are the ends of a diameter.
Answer:We are given the endpoints of a diameter: \( (x_1, y_1) = (3, 4) \) and \( (x_2, y_2) = (2, -7) \). The equation of a circle when the endpoints of a diameter are known is given by: \( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \) Substitute the given coordinates into this formula: \( (x - 3)(x - 2) + (y - 4)(y - (-7)) = 0 \) \( (x - 3)(x - 2) + (y - 4)(y + 7) = 0 \) Now, expand and simplify the terms: \( (x^2 - 2x - 3x + 6) + (y^2 + 7y - 4y - 28) = 0 \) \( (x^2 - 5x + 6) + (y^2 + 3y - 28) = 0 \) Combine like terms:
\( \implies x^2 + y^2 - 5x + 3y + 6 - 28 = 0 \)
\( \implies x^2 + y^2 - 5x + 3y - 22 = 0 \) This is the equation of the circle.In simple words: When you know the two end points of a circle's widest line (diameter), you can directly use a special formula. You just put the x-coordinates and y-coordinates of the two points into \( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \) and then multiply everything out and simplify it. This gives you the circle's equation.

🎯 Exam Tip: Memorize the diameter form of the circle's equation. It saves time and avoids needing to calculate the center (midpoint) and radius separately.

Question 6. Find the equation of the circle through the points (1, 0), (-1, 0) and (0, 1).
Answer:Let the general equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \). The circle passes through the three given points. We will substitute each point into the general equation to form a system of equations. Point 1: \( (1, 0) \) \( (1)^2 + (0)^2 + 2g(1) + 2f(0) + c = 0 \) \( 1 + 0 + 2g + 0 + c = 0 \)
\( \implies 2g + c = -1 \) (Equation 1) Point 2: \( (-1, 0) \) \( (-1)^2 + (0)^2 + 2g(-1) + 2f(0) + c = 0 \) \( 1 + 0 - 2g + 0 + c = 0 \)
\( \implies -2g + c = -1 \) (Equation 2) Point 3: \( (0, 1) \) \( (0)^2 + (1)^2 + 2g(0) + 2f(1) + c = 0 \) \( 0 + 1 + 0 + 2f + c = 0 \)
\( \implies 2f + c = -1 \) (Equation 3) Now, we solve these three equations to find \( g, f, \) and \( c \). Add Equation 1 and Equation 2: \( (2g + c) + (-2g + c) = -1 + (-1) \) \( 2c = -2 \)
\( \implies c = -1 \) Substitute \( c = -1 \) into Equation 1: \( 2g + (-1) = -1 \) \( 2g - 1 = -1 \) \( 2g = 0 \)
\( \implies g = 0 \) Substitute \( c = -1 \) into Equation 3: \( 2f + (-1) = -1 \) \( 2f - 1 = -1 \) \( 2f = 0 \)
\( \implies f = 0 \) Now, substitute the values of \( g = 0, f = 0, \) and \( c = -1 \) into the general equation of the circle: \( x^2 + y^2 + 2(0)x + 2(0)y + (-1) = 0 \) \( x^2 + y^2 + 0x + 0y - 1 = 0 \)
\( \implies x^2 + y^2 - 1 = 0 \) This is the required equation of the circle.In simple words: To find the equation of a circle that goes through three points, we start with the general form of a circle's equation. Then, we substitute each of the three points' x and y values into this equation. This gives us three small equations with three unknowns \( g, f, \) and \( c \). We solve these equations to find the values of \( g, f, \) and \( c \). Finally, we put these values back into the general equation to get the specific circle equation.

🎯 Exam Tip: When finding a circle passing through three points, always use the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \). The process involves setting up and solving a system of three linear equations for \( g, f, \) and \( c \).

Question 7. A circle of area 9π square units has two of its diameters along the lines x + y = 5 and: x - y = 1. Find the equation of the circle.
Answer:The area of the circle is given as \( 9\pi \) square units. The formula for the area of a circle is \( A = \pi r^2 \). So, \( \pi r^2 = 9\pi \)
\( \implies r^2 = 9 \)
\( \implies r = \sqrt{9} = 3 \) (Radius must be positive) Thus, the radius of the circle is 3 units. The intersection point of any two diameters of a circle is its center. We are given two diameter lines: Line 1: \( x + y = 5 \) (Equation 1) Line 2: \( x - y = 1 \) (Equation 2) To find the center \( (h, k) \), we solve this system of equations. Add Equation 1 and Equation 2: \( (x + y) + (x - y) = 5 + 1 \) \( 2x = 6 \)
\( \implies x = \frac{6}{2} = 3 \) Substitute \( x = 3 \) into Equation 1: \( 3 + y = 5 \) \( y = 5 - 3 \)
\( \implies y = 2 \) So, the center of the circle is \( C = (3, 2) \). Now we have the center \( (h, k) = (3, 2) \) and the radius \( r = 3 \). The equation of the circle in standard form is \( (x - h)^2 + (y - k)^2 = r^2 \). Substitute the values: \( (x - 3)^2 + (y - 2)^2 = 3^2 \)
\( \implies (x - 3)^2 + (y - 2)^2 = 9 \) To get the general form, expand the equation: \( (x^2 - 6x + 9) + (y^2 - 4y + 4) = 9 \) \( x^2 + y^2 - 6x - 4y + 13 = 9 \)
\( \implies x^2 + y^2 - 6x - 4y + 4 = 0 \) This is the equation of the circle.In simple words: First, we use the given area of the circle to find its radius. Then, since the center of a circle is where all its diameters cross, we solve the two equations of the diameter lines to find their intersection point, which is the center of the circle. Finally, we use the center and the radius in the standard circle formula \( (x-h)^2 + (y-k)^2 = r^2 \) to get the circle's equation.

🎯 Exam Tip: The intersection of any two diameters always gives the center of the circle. For area-related problems, always find the radius first from the area formula.

Question 8. If \( y = 2\sqrt{2}x + c \) is a tangent to the circle \( x^2 + y^2 = 16 \), find the value of c.
Answer:The given equation of the line is \( y = 2\sqrt{2}x + c \). This is in the form \( y = mx + c \), where \( m = 2\sqrt{2} \). The given equation of the circle is \( x^2 + y^2 = 16 \). This is in the form \( x^2 + y^2 = a^2 \), where \( a^2 = 16 \). So, \( a = 4 \). For a line \( y = mx + c \) to be tangent to a circle \( x^2 + y^2 = a^2 \), the condition is \( c^2 = a^2(1 + m^2) \). Substitute the values of \( a^2 \) and \( m \): \( c^2 = 16(1 + (2\sqrt{2})^2) \) First, calculate \( m^2 \): \( m^2 = (2\sqrt{2})^2 = 2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8 \) Now substitute \( m^2 = 8 \) into the condition: \( c^2 = 16(1 + 8) \) \( c^2 = 16(9) \) \( c^2 = 144 \) To find \( c \), take the square root of both sides: \( c = \pm \sqrt{144} \)
\( \implies c = \pm 12 \) Therefore, the possible values of \( c \) are \( 12 \) and \( -12 \).In simple words: We have a line and a circle, and the line just touches the circle (it's a tangent). There's a special rule that connects the slope of the line (\( m \)), the radius of the circle (\( a \)), and the y-intercept of the line (\( c \)): \( c^2 = a^2(1 + m^2) \). We find \( a^2 \) from the circle's equation and \( m \) from the line's equation, then we plug these numbers into the rule to solve for \( c \).

🎯 Exam Tip: Remember the tangency condition \( c^2 = a^2(1 + m^2) \) for a line \( y = mx + c \) and a circle \( x^2 + y^2 = a^2 \). This formula is crucial for quickly solving such problems without complex calculations.

Question 9. Find the equation of the tangent and normal to the circle \( x^2 + y^2 - 6x + 6y - 8 = 0 \) at (2, 2).
Answer:The equation of the circle is \( x^2 + y^2 - 6x + 6y - 8 = 0 \). This is in the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \). Comparing coefficients, we have: \( 2g = -6 \implies g = -3 \) \( 2f = 6 \implies f = 3 \) \( c = -8 \) The point of tangency is \( (x_1, y_1) = (2, 2) \). **Equation of the Tangent:** The equation of the tangent to the circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) at \( (x_1, y_1) \) is: \( xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 \) Substitute the values of \( x_1, y_1, g, f, \) and \( c \): \( x(2) + y(2) + (-3)(x + 2) + (3)(y + 2) + (-8) = 0 \) \( 2x + 2y - 3x - 6 + 3y + 6 - 8 = 0 \) Combine like terms: \( (2x - 3x) + (2y + 3y) + (-6 + 6 - 8) = 0 \)
\( \implies -x + 5y - 8 = 0 \) This can also be written as \( x - 5y + 8 = 0 \). So, the equation of the tangent is \( x - 5y + 8 = 0 \). **Equation of the Normal:** The normal to the circle at a point is a line perpendicular to the tangent at that point and passes through the center of the circle. It also passes through the point of tangency \( (2, 2) \). From the equation of the tangent \( -x + 5y - 8 = 0 \), the slope of the tangent \( m_t \) can be found by rearranging it to \( 5y = x + 8 \implies y = \frac{1}{5}x + \frac{8}{5} \). So, \( m_t = \frac{1}{5} \). The slope of the normal \( m_n \) will be the negative reciprocal of \( m_t \): \( m_n = -\frac{1}{m_t} = -\frac{1}{1/5} = -5 \). Now use the point-slope form of a line \( y - y_1 = m_n(x - x_1) \) with \( (x_1, y_1) = (2, 2) \) and \( m_n = -5 \): \( y - 2 = -5(x - 2) \) \( y - 2 = -5x + 10 \) \( 5x + y - 2 - 10 = 0 \)
\( \implies 5x + y - 12 = 0 \) Alternatively, the normal passes through the center \( (-g, -f) = (3, -3) \) and the point \( (2, 2) \). Using these two points, we can find the equation of the normal. Slope \( m_n = \frac{2 - (-3)}{2 - 3} = \frac{2+3}{-1} = \frac{5}{-1} = -5 \). Using point \( (2, 2) \) and slope \( -5 \): \( y - 2 = -5(x - 2) \) \( y - 2 = -5x + 10 \) \( 5x + y - 12 = 0 \). Both methods give the same equation for the normal.In simple words: We need to find two lines: a tangent line that just touches the circle at one point, and a normal line that goes through the center and is perpendicular to the tangent. First, we use a special formula for the tangent line, plugging in the given point and the values from the circle's general equation. For the normal line, we find its slope by taking the negative inverse of the tangent's slope, and then use the same point to write its equation.

🎯 Exam Tip: Remember that the tangent formula requires extracting \( g, f, \) and \( c \) from the general equation of the circle. The normal is always perpendicular to the tangent at the point of tangency and passes through the center of the circle.

Question 10. Determine whether the points (- 2, 1), (0, 0) and (- 4, - 3) lie outside, on or inside the circle \( x^2 + y^2 - 5x + 2y - 5 = 0 \).
Answer:Let the equation of the circle be \( S \equiv x^2 + y^2 - 5x + 2y - 5 = 0 \). To determine whether a point \( (x_1, y_1) \) lies outside, on, or inside the circle, we substitute the coordinates into the expression \( S_1 = x_1^2 + y_1^2 - 5x_1 + 2y_1 - 5 \). - If \( S_1 > 0 \), the point lies outside the circle. - If \( S_1 = 0 \), the point lies on the circle. - If \( S_1 < 0 \), the point lies inside the circle. **Point 1: \( (-2, 1) \)** Substitute \( x = -2 \) and \( y = 1 \) into \( S \): \( S_1 = (-2)^2 + (1)^2 - 5(-2) + 2(1) - 5 \) \( S_1 = 4 + 1 + 10 + 2 - 5 \) \( S_1 = 17 - 5 \) \( S_1 = 12 \) Since \( S_1 = 12 > 0 \), the point \( (-2, 1) \) lies **outside** the circle. **Point 2: \( (0, 0) \)** Substitute \( x = 0 \) and \( y = 0 \) into \( S \): \( S_1 = (0)^2 + (0)^2 - 5(0) + 2(0) - 5 \) \( S_1 = 0 + 0 - 0 + 0 - 5 \) \( S_1 = -5 \) Since \( S_1 = -5 < 0 \), the point \( (0, 0) \) lies **inside** the circle. **Point 3: \( (-4, -3) \)** Substitute \( x = -4 \) and \( y = -3 \) into \( S \): \( S_1 = (-4)^2 + (-3)^2 - 5(-4) + 2(-3) - 5 \) \( S_1 = 16 + 9 + 20 - 6 - 5 \) \( S_1 = 45 - 11 \) \( S_1 = 34 \) Since \( S_1 = 34 > 0 \), the point \( (-4, -3) \) lies **outside** the circle.In simple words: To check if a point is inside, outside, or on a circle, we simply plug its x and y values into the circle's equation. If the result is greater than zero, the point is outside. If it's exactly zero, the point is on the circle. If it's less than zero, the point is inside. We do this for each given point.

🎯 Exam Tip: The sign of the expression \( S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c \) directly tells you the position of a point \( (x_1, y_1) \) relative to the circle. This is a fundamental concept in coordinate geometry.

Question 11. Find the centre and radius of the following circles.
(i) \( x^2 + (y + 2)^2 = 0 \)
(ii) \( x^2 + y^2 + 6x - 4y + 4 = 0 \)
(iii) \( x^2 + y^2 - x + 2y - 3 = 0 \)
(iv) \( 2x^2 + 2y^2 - 6x + 4y + 2 = 0 \)
Answer:The general equation of a circle is \( x^2 + y^2 + 2gx + 2fy + c = 0 \). The center of the circle is \( (-g, -f) \) and the radius is \( r = \sqrt{g^2 + f^2 - c} \). (i) **For the circle \( x^2 + (y + 2)^2 = 0 \)** Expand the equation: \( x^2 + y^2 + 4y + 4 = 0 \) Compare with \( x^2 + y^2 + 2gx + 2fy + c = 0 \): \( 2g = 0 \implies g = 0 \) \( 2f = 4 \implies f = 2 \) \( c = 4 \) Center \( (-g, -f) = (-(0), -(2)) = (0, -2) \) Radius \( r = \sqrt{g^2 + f^2 - c} = \sqrt{0^2 + 2^2 - 4} = \sqrt{0 + 4 - 4} = \sqrt{0} = 0 \) This equation represents a point circle (a circle with zero radius), which is just the point \( (0, -2) \). (ii) **For the circle \( x^2 + y^2 + 6x - 4y + 4 = 0 \)** Compare with \( x^2 + y^2 + 2gx + 2fy + c = 0 \): \( 2g = 6 \implies g = 3 \) \( 2f = -4 \implies f = -2 \) \( c = 4 \) Center \( (-g, -f) = (-(3), -(-2)) = (-3, 2) \) Radius \( r = \sqrt{g^2 + f^2 - c} = \sqrt{3^2 + (-2)^2 - 4} = \sqrt{9 + 4 - 4} = \sqrt{9} = 3 \) (iii) **For the circle \( x^2 + y^2 - x + 2y - 3 = 0 \)** Compare with \( x^2 + y^2 + 2gx + 2fy + c = 0 \): \( 2g = -1 \implies g = -\frac{1}{2} \) \( 2f = 2 \implies f = 1 \) \( c = -3 \) Center \( (-g, -f) = (-(-\frac{1}{2}), -(1)) = (\frac{1}{2}, -1) \) Radius \( r = \sqrt{g^2 + f^2 - c} = \sqrt{(-\frac{1}{2})^2 + (1)^2 - (-3)} \) \( r = \sqrt{\frac{1}{4} + 1 + 3} = \sqrt{\frac{1}{4} + 4} = \sqrt{\frac{1 + 16}{4}} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \) (iv) **For the circle \( 2x^2 + 2y^2 - 6x + 4y + 2 = 0 \)** First, divide the entire equation by 2 to get it into the standard general form where the coefficients of \( x^2 \) and \( y^2 \) are 1: \( \frac{2x^2}{2} + \frac{2y^2}{2} - \frac{6x}{2} + \frac{4y}{2} + \frac{2}{2} = 0 \)
\( \implies x^2 + y^2 - 3x + 2y + 1 = 0 \) Compare with \( x^2 + y^2 + 2gx + 2fy + c = 0 \): \( 2g = -3 \implies g = -\frac{3}{2} \) \( 2f = 2 \implies f = 1 \) \( c = 1 \) Center \( (-g, -f) = (-(-\frac{3}{2}), -(1)) = (\frac{3}{2}, -1) \) Radius \( r = \sqrt{g^2 + f^2 - c} = \sqrt{(-\frac{3}{2})^2 + (1)^2 - 1} \) \( r = \sqrt{\frac{9}{4} + 1 - 1} = \sqrt{\frac{9}{4}} = \frac{3}{2} \)In simple words: For each given circle equation, we compare it to the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \). From this comparison, we find the values of \( g, f, \) and \( c \). The center of the circle is always \( (-g, -f) \), and the radius is found using the formula \( \sqrt{g^2 + f^2 - c} \). If the \( x^2 \) and \( y^2 \) terms have coefficients other than 1, divide the whole equation by that coefficient first.

🎯 Exam Tip: Always normalize the circle equation (make coefficients of \( x^2 \) and \( y^2 \) equal to 1) before extracting \( g, f, \) and \( c \). Be careful with signs when determining \( -g \) and \( -f \), especially with negative values.

Question 12. If the equation \( 3x^2 + (3 - p) xy + qy^2 - 2px = 8pq \) represents a circle, find p and q. Also determine the centre and radius of the circle.
Answer:The given equation is \( 3x^2 + (3 - p) xy + qy^2 - 2px = 8pq \). For this equation to represent a circle, two conditions must be met: 1. The coefficient of \( xy \) must be zero. 2. The coefficient of \( x^2 \) must be equal to the coefficient of \( y^2 \). From condition 1: The coefficient of \( xy \) is \( (3 - p) \). So, \( 3 - p = 0 \)
\( \implies p = 3 \) From condition 2: The coefficient of \( x^2 \) is \( 3 \), and the coefficient of \( y^2 \) is \( q \). So, \( q = 3 \) Now substitute \( p = 3 \) and \( q = 3 \) back into the original equation: \( 3x^2 + (3 - 3)xy + 3y^2 - 2(3)x = 8(3)(3) \) \( 3x^2 + 0xy + 3y^2 - 6x = 72 \) \( 3x^2 + 3y^2 - 6x = 72 \) To find the center and radius, we need to bring this equation into the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \). Divide the entire equation by 3: \( \frac{3x^2}{3} + \frac{3y^2}{3} - \frac{6x}{3} = \frac{72}{3} \)
\( \implies x^2 + y^2 - 2x = 24 \) Rearrange to match the general form: \( x^2 + y^2 - 2x - 24 = 0 \) Now, compare with \( x^2 + y^2 + 2gx + 2fy + c = 0 \): \( 2g = -2 \implies g = -1 \) \( 2f = 0 \implies f = 0 \) (since there is no \( y \) term) \( c = -24 \) Center of the circle \( (-g, -f) = (-( -1 ), -(0)) = (1, 0) \) Radius of the circle \( r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (0)^2 - (-24)} \) \( r = \sqrt{1 + 0 + 24} = \sqrt{25} = 5 \) So, \( p = 3 \), \( q = 3 \). The center of the circle is \( (1, 0) \) and the radius is \( 5 \).In simple words: For an equation to be a circle, there can't be an \( xy \) term, and the numbers in front of \( x^2 \) and \( y^2 \) must be the same. We use these two rules to find the values of \( p \) and \( q \). Once we have \( p \) and \( q \), we plug them back into the equation, simplify it to the standard general form, and then find the center and radius using their respective formulas.

🎯 Exam Tip: Always remember the two key conditions for a second-degree equation to represent a circle: coefficient of \( xy \) is zero, and coefficients of \( x^2 \) and \( y^2 \) are equal. Apply these conditions first to find unknown parameters.