Topic 1: Standard & Simplest Equations

1. Given: Center $(0,0)$, Radius $r=5$.
   Formula: $x^2 + y^2 = r^2$
   Solution: $x^2 + y^2 = 5^2$ $\Rightarrow$ $x^2 + y^2 = 25$.
   Ans: $x^2 + y^2 = 25$
2. Given: Center $(h,k) = (2, -3)$, Radius $r=4$.
   Formula: $(x-h)^2 + (y-k)^2 = r^2$
   Solution: $(x-2)^2 + (y - (-3))^2 = 4^2$ $\Rightarrow$ $(x-2)^2 + (y+3)^2 = 16$.
   Ans: $(x-2)^2 + (y+3)^2 = 16$
3. Equation: $(x - 5)^2 + (y + 2)^2 = 36$.
   Compare with $(x-h)^2 + (y-k)^2 = r^2$.
   $h=5$, $k=-2$, $r^2=36$ $\Rightarrow$ $r=6$.
   Ans: Center $(5, -2)$, Radius $6$
4. Passes through origin $(0,0)$, radius $r=3$. Center on positive x-axis means $k=0$ and $h > 0$.
   Since it passes through origin, distance from Center $(h,0)$ to $(0,0)$ is radius.
   $h = 3$. Center is $(3,0)$.
   Equation: $(x-3)^2 + y^2 = 3^2$.
   Ans: $(x-3)^2 + y^2 = 9$
5. Center $C(-1, 2)$, passes through $P(3, 5)$.
   Radius $r = CP$ $= \sqrt{(3 - (-1))^2 + (5 - 2)^2}$ $= \sqrt{4^2 + 3^2}$ $= \sqrt{16+9}$ $= \sqrt{25}$ $= 5$.
   Equation: $(x+1)^2 + (y-2)^2 = 5^2$.
   Ans: $(x+1)^2 + (y-2)^2 = 25$
6. Center $C(3, 4)$ touches y-axis.
   Radius = perpendicular distance from Center to y-axis $= |x\text{-coordinate}|$ $= 3$.
   Equation: $(x-3)^2 + (y-4)^2 = 3^2$.
   Ans: $(x-3)^2 + (y-4)^2 = 9$
7. Center $(0,0)$, Radius $r=\sqrt{5}$.
   Equation: $x^2 + y^2 = (\sqrt{5})^2$.
   Ans: $x^2 + y^2 = 5$
8. Equation: $x^2 + y^2 = 25$. Point $(3, 4)$.
   Check LHS: $3^2 + 4^2$ $= 9 + 16$ $= 25$.
   Since LHS = RHS, point lies on circle.
   Ans: Yes

Topic 2: General Equation (Center & Radius)

1. $x^2 + y^2 - 4x - 8y - 45 = 0$.
   $2g = -4$ $\Rightarrow$ $g = -2$. $2f = -8$ $\Rightarrow$ $f = -4$. $c = -45$.
   Center $(-g, -f) = (2, 4)$.
   Radius $r = \sqrt{g^2 + f^2 - c}$ $= \sqrt{(-2)^2 + (-4)^2 - (-45)}$ $= \sqrt{4 + 16 + 45}$ $= \sqrt{65}$.
   Ans: C$(2, 4)$, $r=\sqrt{65}$
2. $x^2 + y^2 + 8x + 10y - 8 = 0$.
   $2g = 8$ $\Rightarrow$ $g = 4$. $2f = 10$ $\Rightarrow$ $f = 5$. $c = -8$.
   Center $(-g, -f) = (-4, -5)$.
   Radius $r = \sqrt{16 + 25 - (-8)}$ $= \sqrt{16 + 25 + 8}$ $= \sqrt{49}$ $= 7$.
   Ans: C$(-4, -5)$, $r=7$
3. $2x^2 + 2y^2 - x = 0$. Divide by 2: $x^2 + y^2 - \frac{1}{2}x = 0$.
   $2g = -1/2$ $\Rightarrow$ $g = -1/4$. $2f = 0$ $\Rightarrow$ $f = 0$. $c=0$.
   Center $(1/4, 0)$. Radius $r = \sqrt{(1/4)^2 + 0 - 0}$ $= 1/4$.
   Ans: C$(0.25, 0)$, $r=0.25$
4. Points $(1,0), (-1,0), (0,1)$.
   $(1,0)$ and $(-1,0)$ are equidistant from y-axis, center lies on $x=0$.
   Let equation be $x^2+y^2+2fy+c=0$ (since center on y-axis, $g=0$).
   Passes $(1,0)$ $\Rightarrow$ $1+c=0$ $\Rightarrow$ $c=-1$.
   Passes $(0,1)$ $\Rightarrow$ $1+2f+c=0$ $\Rightarrow$ $1+2f-1=0$ $\Rightarrow$ $f=0$.
   Center $(0,0)$, $c=-1$. Eq: $x^2+y^2-1=0$.
   Ans: $x^2 + y^2 = 1$
5. Old Circle: $x^2 + y^2 - 6x + 12y + 15 = 0$. Center $(3, -6)$.
   Old Radius $r = \sqrt{9 + 36 - 15}$ $= \sqrt{30}$. Area $A = 30\pi$.
   New Circle is concentric $\Rightarrow$ Center $(3, -6)$.
   New Area $= 2 \times 30\pi$ $= 60\pi$ $\Rightarrow$ $R^2 = 60$.
   Equation: $(x-3)^2 + (y+6)^2 = 60$.
   Ans: $(x-3)^2 + (y+6)^2 = 60$
6. $x^2 + y^2 + 4x - 6y + 13 = 0$.
   Radius $r = \sqrt{g^2+f^2-c}$ $= \sqrt{(2)^2 + (-3)^2 - 13}$ $= \sqrt{4 + 9 - 13}$ $= \sqrt{0}$ $= 0$.
   Since radius is 0, it is a point circle.
   Ans: Proved
7. For second degree equation to be a circle, coefficient of $x^2$ must equal coefficient of $y^2$.
   $k = 1$.
   Ans: $k=1$
8. Passes through origin $(0,0)$.
   Substitute $x=0, y=0$ into equation: $0^2 + 0^2 + 0 + 0 + c = 0$ $\Rightarrow$ $c=0$.
   Ans: $c=0$
9. General form passing through $(0,0) \Rightarrow c=0$.
   Passes through $(a,0)$ $\Rightarrow$ $a^2 + 2ga = 0$ $\Rightarrow$ $g = -a/2$.
   Passes through $(0,b)$ $\Rightarrow$ $b^2 + 2fb = 0$ $\Rightarrow$ $f = -b/2$.
   Eq: $x^2 + y^2 - ax - by = 0$.
   Ans: $x^2 + y^2 - ax - by = 0$
10. $3x^2 + 3y^2 - 12x + 15y - 23 = 0$. First divide by 3.
    $x^2 + y^2 - 4x + 5y - \frac{23}{3} = 0$.
    $2g = -4$ $\Rightarrow$ $g = -2$. $2f = 5$ $\Rightarrow$ $f = 5/2$.
    Center $(-g, -f) = (2, -5/2)$.
    Ans: C$(2, -2.5)$

Topic 3: Nature of Circle

1. $x^2 + y^2 - 2x + 4y + 10 = 0$.
   $g=-1, f=2, c=10$.
   $r^2 = g^2 + f^2 - c$ $= 1 + 4 - 10$ $= -5$.
   Since $r^2 < 0$, radius is imaginary.
   Ans: No (Imaginary Circle)
2. $g=1, f=-3, c=10$.
   $r^2 = 1^2 + (-3)^2 - 10$ $= 1 + 9 - 10$ $= 0$.
   Radius is zero.
   Ans: Point Circle
3. For real circle, $r^2 > 0$ $\Rightarrow$ $g^2 + f^2 - c > 0$.
   $g=-2, f=-1$.
   $(-2)^2 + (-1)^2 - c > 0$ $\Rightarrow$ $4 + 1 - c > 0$ $\Rightarrow$ $5 - c > 0$ $\Rightarrow$ $c < 5$.
   Ans: $c < 5$
4. $g=1, f=1, c=5$.
   $r^2 = 1^2 + 1^2 - 5$ $= 2 - 5$ $= -3$.
   Negative value under root implies imaginary.
   Ans: Proved
5. $g=-2, f=-2, c=8$.
   $r = \sqrt{4 + 4 - 8}$ $= \sqrt{0}$ $= 0$.
   It implies the circle has shrunk to a single point $(2,2)$.
   Ans: $r=0$ (Point Circle)

Topic 4: Diameter Form

1. Endpoints $(1, 2)$ and $(3, -4)$.
   Formula: $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
   $(x-1)(x-3) + (y-2)(y+4) = 0$.
   $x^2 - 4x + 3 + y^2 + 2y - 8 = 0$ $\Rightarrow$ $x^2 + y^2 - 4x + 2y - 5 = 0$.
   Ans: $x^2 + y^2 - 4x + 2y - 5 = 0$
2. Endpoints $(-2, 3)$ and $(-3, 5)$.
   $(x+2)(x+3) + (y-3)(y-5) = 0$.
   Ans: $(x+2)(x+3) + (y-3)(y-5) = 0$
3. Endpoints $(0,0)$ and $(4,4)$.
   $(x-0)(x-4) + (y-0)(y-4) = 0$ $\Rightarrow$ $x(x-4) + y(y-4) = 0$.
   Ans: $x^2 + y^2 - 4x - 4y = 0$
4. Line $4x + 3y = 24$.
   At x-axis ($y=0$), $4x=24 \Rightarrow x=6$. Point A$(6,0)$.
   At y-axis ($x=0$), $3y=24 \Rightarrow y=8$. Point B$(0,8)$.
   Diameter ends $(6,0)$ and $(0,8)$.
   $(x-6)(x-0) + (y-0)(y-8) = 0$.
   Ans: $x^2 + y^2 - 6x - 8y = 0$
5. Circle center from eq: $2g=-4 \to g=-2 \Rightarrow C(2, 3)$.
   One end $A(3,4)$. Let other end be $B(x,y)$.
   Center is midpoint: $\frac{3+x}{2}=2 \Rightarrow x=1$. $\frac{4+y}{2}=3 \Rightarrow y=2$.
   Ans: $(1, 2)$
6. Ends $(0,0)$ and $(2,2)$.
   $(x-0)(x-2) + (y-0)(y-2) = 0$.
   Ans: $x^2 + y^2 - 2x - 2y = 0$
7. In rectangle ABCD, diagonal AC is diameter.
   Ends A$(1,2)$ and C$(5,6)$.
   $(x-1)(x-5) + (y-2)(y-6) = 0$.
   Ans: $x^2 + y^2 - 6x - 8y + 17 = 0$

Topic 5: Parametric Equations

1. $x^2 + y^2 = 16 \Rightarrow r=4$.
   $x = r\cos\theta$, $y = r\sin\theta$.
   Ans: $x=4\cos\theta, y=4\sin\theta$
2. Center $(1, -2)$, Radius $r=5$.
   $x = h + r\cos\theta$ $\Rightarrow$ $x = 1 + 5\cos\theta$.
   $y = k + r\sin\theta$ $\Rightarrow$ $y = -2 + 5\sin\theta$.
   Ans: $x=1+5\cos\theta, y=-2+5\sin\theta$
3. Given $x-3 = 2\cos\theta$ and $y-4 = 2\sin\theta$.
   Square and add: $(x-3)^2 + (y-4)^2$ $= 4(\cos^2\theta + \sin^2\theta)$ $= 4$.
   Ans: $(x-3)^2 + (y-4)^2 = 4$
4. Compare with standard parametric form.
   $h=-1$, $k=2$, $r=3$.
   Ans: C$(-1, 2)$, $r=3$
5. $x^2 + y^2 + 4x - 6y - 12 = 0$.
   $C(-2, 3)$. $r = \sqrt{4+9-(-12)}$ $= \sqrt{25}$ $= 5$.
   Parametric: $x = -2 + 5\cos\theta$, $y = 3 + 5\sin\theta$.
   Ans: $x=-2+5\cos\theta, y=3+5\sin\theta$

Topic 6: Position of a Point

1. $S_1 = x^2 + y^2 - 9$. Point $(2,1)$.
   $S_1 = 2^2 + 1^2 - 9$ $= 4 + 1 - 9$ $= -4$.
   Since $S_1 < 0$, point is inside.
   Ans: Inside
2. $S_1 = 1^2 + (-2)^2 - 4(1) + 2(-2) - 11$.
   $S_1 = 1 + 4 - 4 - 4 - 11$ $= -14 < 0$.
   Ans: Inside
3. Point on circle $\Rightarrow S_1 = 0$.
   $1^2 + 2^2 - 2(1) + 6(2) + k = 0$.
   $1 + 4 - 2 + 12 + k = 0$ $\Rightarrow$ $15 + k = 0$ $\Rightarrow$ $k = -15$.
   Ans: $k=-15$
4. $S_1 = (-3)^2 + (-4)^2 - 20$ $= 9 + 16 - 20$ $= 5$.
   $S_1 > 0$.
   Ans: Outside
5. Inside $\Rightarrow S_1 < 0$. Circle: $x^2+y^2-8=0$.
   $a^2 + a^2 - 8 < 0$ $\Rightarrow$ $2a^2 < 8$ $\Rightarrow$ $a^2 < 4$.
   $-2 < a < 2$.
   Ans: $a \in (-2, 2)$
6. Center of 1st circle: $(-g, -f) = (3, 4)$.
   Check $(3,4)$ in 2nd circle $x^2 + y^2 - 1 = 0$.
   $3^2 + 4^2 - 1$ $= 9 + 16 - 1$ $= 24 > 0$.
   Ans: No (Outside)
7. Power $S_1 = 1^2 + 1^2 - 4(1) + 6(1) - 3$.
   $S_1 = 1 + 1 - 4 + 6 - 3$ $= 1$.
   Ans: 1

Topic 7: Intercepts & Line Interactions

1. X-intercept $= 2\sqrt{g^2 - c}$.
   $g=4, c=-5$.
   Len $= 2\sqrt{4^2 - (-5)}$ $= 2\sqrt{16+5}$ $= 2\sqrt{21}$.
   Ans: $2\sqrt{21}$
2. Y-intercept $= 2\sqrt{f^2 - c}$.
   $2f=6$ $\Rightarrow$ $f=3$, $c=4$.
   Len $= 2\sqrt{3^2 - 4}$ $= 2\sqrt{5}$.
   Ans: $2\sqrt{5}$
3. Condition to cut X-axis: $g^2 > c$.
   $g=1, c=5$. $1^2 = 1$. Since $1 < 5$, it does not cut.
   Ans: No
4. Touches x-axis $\Rightarrow$ X-intercept is 0.
   $2\sqrt{g^2 - c} = 0$ $\Rightarrow$ $g^2 = c$.
   Ans: $g^2 = c$
5. Circle $x^2 + y^2 = 25$. Line $y = 3$.
   Intersection: $x^2 + 3^2 = 25$ $\Rightarrow$ $x^2 = 16$ $\Rightarrow$ $x = \pm 4$.
   Points are $(-4,3)$ and $(4,3)$. Distance $= 4 - (-4) = 8$.
   Ans: 8 units
6. Center $(0,0)$, Radius $r=5$. Line $3x+4y-25=0$.
   Dist $d = \frac{|3(0) + 4(0) - 25|}{\sqrt{3^2+4^2}}$ $= \frac{25}{5}$ $= 5$.
   Since $d = r$, it is a Tangent.
   Ans: Tangent
7. Pass origin $\Rightarrow c=0$. Center $(3,4) \Rightarrow g=-3$.
   X-int $= 2\sqrt{g^2 - c}$ $= 2\sqrt{(-3)^2 - 0}$ $= 2(3)$ $= 6$.
   Ans: 6 units
8. $2f = -13$ $\Rightarrow$ $f = -6.5$. $c = 30$.
   Y-int $= 2\sqrt{(-6.5)^2 - 30}$ $= 2\sqrt{42.25 - 30}$ $= 2\sqrt{12.25}$.
   $2(3.5) = 7$.
   Ans: 7 units