Class 12 FSc Math Chapter 6 Conic Sections Exercise 6.1 Solved PDF Notes | FBISE & BISE Punjab Boards (New 2025-2026 Syllabus)
Exercise 6.1 of Class 12 Mathematics (Chapter 6: Conic Sections) focuses on the analytical geometry of the Circle. A circle is defined as the locus of a point $P(x, y)$ moving in a plane such that its distance from a fixed point $C(h, k)$, known as the center, remains constant. This constant distance $r$ is called the radius of the circle.
Using the distance formula between the moving point $P(x, y)$ and the fixed center $C(h, k)$:
$$\sqrt{(x - h)^2 + (y - k)^2} = r$$Squaring both sides yields the Standard Form Equation of a Circle:
$$(x - h)^2 + (y - k)^2 = r^2$$When the center is located at the origin $C(0, 0)$, the equation reduces to the simplest standard form:
$$x^2 + y^2 = r^2$$Expanding the standard equation $(x - h)^2 + (y - k)^2 = r^2$ gives:
$$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = 0$$By defining $g = -h$, $f = -k$, and $c = h^2 + k^2 - r^2$, we obtain the General Form Equation of a Circle:
$$x^2 + y^2 + 2gx + 2fy + c = 0$$For any equation given in general form, the center and radius are determined as follows:
An equation of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ represents a circle if and only if:
| Form / Property | Mathematical Formula / Equation | Center Coordinates | Radius Formula |
|---|---|---|---|
| Central Form (Origin) | $x^2 + y^2 = r^2$ | $(0, 0)$ | $r$ |
| Standard Form | $(x - h)^2 + (y - k)^2 = r^2$ | $(h, k)$ | $r$ |
| General Form | $x^2 + y^2 + 2gx + 2fy + c = 0$ | $(-g, -f)$ | $r = \sqrt{g^2 + f^2 - c}$ |
| Diameter Form | $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ | $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ | $r = \frac{1}{2}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ |
| Touch x-axis Condition | $x^2 + y^2 + 2gx + 2fy + g^2 = 0$ | $(-g, -f)$ | $r = |f|$ |
| Touch y-axis Condition | $x^2 + y^2 + 2gx + 2fy + f^2 = 0$ | $(-g, -f)$ | $r = |g|$ |
(a) Center $C(0, 0)$ and radius $r = 5$.
Solution:
Using the standard equation of a circle centered at origin $(0,0)$:
$$x^2 + y^2 = r^2$$Substitute $r = 5$ into the equation:
$$x^2 + y^2 = (5)^2$$ $$x^2 + y^2 = 25$$Final Answer: $x^2 + y^2 - 25 = 0$
(b) Center $C(-3, 5)$ and radius $r = \sqrt{7}$.
Solution:
Using the standard equation of a circle with center $(h, k) = (-3, 5)$ and radius $r = \sqrt{7}$:
$$(x - h)^2 + (y - k)^2 = r^2$$Substitute $h = -3$, $k = 5$, and $r = \sqrt{7}$:
$$(x - (-3))^2 + (y - 5)^2 = (\sqrt{7})^2$$ $$(x + 3)^2 + (y - 5)^2 = 7$$Expanding the binomial terms:
$$(x^2 + 6x + 9) + (y^2 - 10y + 25) = 7$$ $$x^2 + y^2 + 6x - 10y + 34 - 7 = 0$$ $$x^2 + y^2 + 6x - 10y + 27 = 0$$Final Answer: $x^2 + y^2 + 6x - 10y + 27 = 0$
(c) Ends of a diameter at $A(-3, 2)$ and $B(5, -6)$.
Solution:
Method 1: Using Diameter Equation Form
The equation of a circle with endpoints of diameter $(x_1, y_1) = (-3, 2)$ and $(x_2, y_2) = (5, -6)$ is:
$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$Substitute the given coordinates:
$$(x - (-3))(x - 5) + (y - 2)(y - (-6)) = 0$$ $$(x + 3)(x - 5) + (y - 2)(y + 6) = 0$$Expand both products:
$$(x^2 - 5x + 3x - 15) + (y^2 + 6y - 2y - 12) = 0$$ $$(x^2 - 2x - 15) + (y^2 + 4y - 12) = 0$$ $$x^2 + y^2 - 2x + 4y - 27 = 0$$Final Answer: $x^2 + y^2 - 2x + 4y - 27 = 0$
(a) $x^2 + y^2 - 6x + 4y - 12 = 0$
Solution:
Compare the given equation with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$:
$$2g = -6 \implies g = -3$$ $$2f = 4 \implies f = 2$$ $$c = -12$$Center:
$$\text{Center } C = (-g, -f) = (-(-3), -(2)) = (3, -2)$$Radius:
$$r = \sqrt{g^2 + f^2 - c}$$ $$r = \sqrt{(-3)^2 + (2)^2 - (-12)}$$ $$r = \sqrt{9 + 4 + 12} = \sqrt{25} = 5 \text{ units}$$Final Answer: Center $= (3, -2)$, Radius $= 5$
(b) $2x^2 + 2y^2 - 8x + 12y + 1 = 0$
Solution:
First, reduce the equation to general form by dividing all terms by $2$:
$$x^2 + y^2 - 4x + 6y + \frac{1}{2} = 0$$Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$:
$$2g = -4 \implies g = -2$$ $$2f = 6 \implies f = 3$$ $$c = \frac{1}{2}$$Center:
$$\text{Center } C = (-g, -f) = (2, -3)$$Radius:
$$r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (3)^2 - \frac{1}{2}}$$ $$r = \sqrt{4 + 9 - \frac{1}{2}} = \sqrt{13 - \frac{1}{2}} = \sqrt{\frac{26 - 1}{2}} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} \text{ units}$$Final Answer: Center $= (2, -3)$, Radius $= \frac{5}{\sqrt{2}}$
Solution:
Let the general equation of the circle be:
$$x^2 + y^2 + 2gx + 2fy + c = 0 \quad \text{--- (Equation 1)}$$Step 1: Substitute point $A(1, 2)$ into Equation 1:
$$(1)^2 + (2)^2 + 2g(1) + 2f(2) + c = 0$$ $$1 + 4 + 2g + 4f + c = 0$$ $$2g + 4f + c = -5 \quad \text{--- (Equation 2)}$$Step 2: Substitute point $B(3, -4)$ into Equation 1:
$$(3)^2 + (-4)^2 + 2g(3) + 2f(-4) + c = 0$$ $$9 + 16 + 6g - 8f + c = 0$$ $$6g - 8f + c = -25 \quad \text{--- (Equation 3)}$$Step 3: Substitute point $C(5, -6)$ into Equation 1:
$$(5)^2 + (-6)^2 + 2g(5) + 2f(-6) + c = 0$$ $$25 + 36 + 10g - 12f + c = 0$$ $$10g - 12f + c = -61 \quad \text{--- (Equation 4)}$$Step 4: Eliminate constant $c$ by subtracting equations:
Subtract Eq. 2 from Eq. 3:
$$(6g - 8f + c) - (2g + 4f + c) = -25 - (-5)$$ $$4g - 12f = -20$$Divide by 4:
$$g - 3f = -5 \quad \text{--- (Equation 5)}$$Subtract Eq. 3 from Eq. 4:
$$(10g - 12f + c) - (6g - 8f + c) = -61 - (-25)$$ $$4g - 4f = -36$$Divide by 4:
$$g - f = -9 \quad \text{--- (Equation 6)}$$Step 5: Solve for $g$ and $f$:
Subtract Eq. 5 from Eq. 6:
$$(g - f) - (g - 3f) = -9 - (-5)$$ $$2f = -4 \implies f = -2$$Substitute $f = -2$ into Eq. 6:
$$g - (-2) = -9 \implies g + 2 = -9 \implies g = -11$$Step 6: Solve for $c$:
Substitute $g = -11$ and $f = -2$ into Eq. 2:
$$2(-11) + 4(-2) + c = -5$$ $$-22 - 8 + c = -5$$ $$-30 + c = -5 \implies c = 25$$Step 7: Substitute $g, f, c$ into Equation 1:
$$x^2 + y^2 + 2(-11)x + 2(-2)y + 25 = 0$$ $$x^2 + y^2 - 22x - 4y + 25 = 0$$Final Answer: $x^2 + y^2 - 22x - 4y + 25 = 0$
Solution:
Let the general equation of the circle be:
$$x^2 + y^2 + 2gx + 2fy + c = 0 \quad \text{--- (Equation 1)}$$The center of this circle is $C(-g, -f)$.
Step 1: Apply condition that center $C(-g, -f)$ lies on $3x + 4y - 7 = 0$:
$$3(-g) + 4(-f) - 7 = 0$$ $$-3g - 4f - 7 = 0 \implies 3g + 4f = -7 \quad \text{--- (Equation 2)}$$Step 2: Circle passes through $A(1, 2)$:
$$(1)^2 + (2)^2 + 2g(1) + 2f(2) + c = 0$$ $$5 + 2g + 4f + c = 0 \implies 2g + 4f + c = -5 \quad \text{--- (Equation 3)}$$Step 3: Circle passes through $B(4, 3)$:
$$(4)^2 + (3)^2 + 2g(4) + 2f(3) + c = 0$$ $$16 + 9 + 8g + 6f + c = 0$$ $$8g + 6f + c = -25 \quad \text{--- (Equation 4)}$$Step 4: Subtract Eq. 3 from Eq. 4 to eliminate $c$:
$$(8g + 6f + c) - (2g + 4f + c) = -25 - (-5)$$ $$6g + 2f = -20$$Divide by 2:
$$3g + f = -10 \quad \text{--- (Equation 5)}$$Step 5: Solve simultaneous linear equations Eq. 2 and Eq. 5:
Subtract Eq. 5 from Eq. 2:
$$(3g + 4f) - (3g + f) = -7 - (-10)$$ $$3f = 3 \implies f = 1$$Substitute $f = 1$ into Eq. 5:
$$3g + 1 = -10 \implies 3g = -11 \implies g = -\frac{11}{3}$$Step 6: Find $c$ using Eq. 3:
$$2\left(-\frac{11}{3}\right) + 4(1) + c = -5$$ $$-\frac{22}{3} + 4 + c = -5$$ $$c = -5 - 4 + \frac{22}{3} = -9 + \frac{22}{3} = \frac{-27 + 22}{3} = -\frac{5}{3}$$Step 7: Substitute parameters into general equation:
$$x^2 + y^2 + 2\left(-\frac{11}{3}\right)x + 2(1)y - \frac{5}{3} = 0$$ $$x^2 + y^2 - \frac{22}{3}x + 2y - \frac{5}{3} = 0$$Multiply the entire equation by 3 to clear fractions:
$$3x^2 + 3y^2 - 22x + 6y - 5 = 0$$Final Answer: $3x^2 + 3y^2 - 22x + 6y - 5 = 0$
(i) the x-axis if $g^2 = c$
(ii) the y-axis if $f^2 = c$
Proof:
Case (i): Circle touching x-axis
The equation of the x-axis is $y = 0$. Substituting $y = 0$ into the circle equation yields:
$$x^2 + (0)^2 + 2gx + 2f(0) + c = 0$$ $$x^2 + 2gx + c = 0$$This is a quadratic equation in $x$. For the circle to be tangent to (touch) the x-axis, this quadratic equation must have exactly one real root (equal roots). Therefore, its discriminant ($\Delta$) must equal zero:
$$\Delta = B^2 - 4AC = 0$$Here $A = 1, B = 2g, C = c$:
$$(2g)^2 - 4(1)(c) = 0$$ $$4g^2 - 4c = 0$$ $$4(g^2 - c) = 0 \implies g^2 = c$$(Hence Proved)
Case (ii): Circle touching y-axis
The equation of the y-axis is $x = 0$. Substituting $x = 0$ into the circle equation yields:
$$(0)^2 + y^2 + 2g(0) + 2fy + c = 0$$ $$y^2 + 2fy + c = 0$$For the circle to touch the y-axis, this quadratic in $y$ must have equal roots ($\Delta = 0$):
$$\Delta = (2f)^2 - 4(1)(c) = 0$$ $$4f^2 - 4c = 0$$ $$4(f^2 - c) = 0 \implies f^2 = c$$(Hence Proved)
Q1: What is the center of the circle $x^2 + y^2 + 8x - 10y + 16 = 0$?
Q2: What is the radius of the circle $3x^2 + 3y^2 - 12x + 18y - 9 = 0$?
Q3: If a circle touches both coordinate axes, which condition must hold true?
Q4: A circle has radius $r = 0$. What geometric figure does it represent?
Q5: What is the equation of a circle centered at $(-2, 3)$ passing through the origin $(0, 0)$?
Q6: Which equation represents the diameter form of a circle with endpoints $(x_1, y_1)$ and $(x_2, y_2)$?
Q7: If $g^2 + f^2 - c < 0$, the circle is classified as:
Q8: What is the length of the intercept made by the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ on the x-axis?
Q1: Why do students lose marks when finding the center from general equations like $2x^2 + 2y^2 - 8x + 12y + 1 = 0$?
Ans: The most common mistake is directly taking $2g = -8$ and $2f = 12$ without first dividing the entire equation by the coefficient of $x^2$ and $y^2$ (which is $2$). The standard general form formula $C(-g, -f)$ applies only when the coefficients of $x^2$ and $y^2$ are strictly equal to $1$. Always divide by the leading coefficient first before comparing coefficients.
Q2: What is the difference between $g, f$ values and center coordinates $(h, k)$?
Ans: Students often confuse the sign of $g$ and $f$ with center coordinates. In general form $x^2 + y^2 + 2gx + 2fy + c = 0$, the center is $(-g, -f)$, not $(g, f)$. If $2g = -6 \implies g = -3$, then the x-coordinate of the center is $-g = -(-3) = +3$. Neglecting this sign flip is a frequent cause of mark deduction in FBISE and BISE board exams.
Q3: How do I choose between standard form $(x-h)^2 + (y-k)^2 = r^2$ and general form $x^2 + y^2 + 2gx + 2fy + c = 0$ when solving exam questions?
Ans: Use standard form when you are explicitly given (or can easily find) the center $(h, k)$ and radius $r$. Use general form when given conditions involving three unknown parameters (like three points passing through the circle, or points and line equations containing the center), as it yields linear equations in $g, f$, and $c$ that are simple to solve simultaneously.