Questions & Answers: "Circles"

Let the first circle be $S_1: x^2+y^2=4$. Its center is $C_1=(0,0)$ and its radius is $R_1=2$. Let the second circle be $S_2: x^2+y^2-6x-8y=24$. Its center is $C_2=(-(-6)/2, -(-8)/2) = (3,4)$. Its radius is $R_2 = \sqrt{3^2+4^2 - (-24)} = \sqrt{9+16+24} = \sqrt{49}=7$. The distance between the centers $d = C_1C_2 = \sqrt{(3-0)^2+(4-0)^2} = \sqrt{9+16} = \sqrt{25}=5$. Now, we compare $d$ with $R_1+R_2$ and $|R_1-R_2|$. $R_1+R_2 = 2+7 = 9$. $|R_1-R_2| = |2-7| = 5$. Since $d = |R_1-R_2|$ (i.e., $5=5$), the two circles touch each other internally. When two circles touch internally, they have exactly one common tangent.

The equation of a family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$. So, the equation of the required circle is $(x^2+y^2-8x-2y+7) + \lambda(x^2+y^2-4x+2y-1) = 0$. This circle passes through the point $(1,2)$. Substituting $x=1$ and $y=2$ into the equation: $(1^2+2^2-8(1)-2(2)+7) + \lambda(1^2+2^2-4(1)+2(2)-1) = 0$ $(1+4-8-4+7) + \lambda(1+4-4+4-1) = 0$ $(12-12) + \lambda(9-5) = 0$ $0 + 4\lambda = 0 \Rightarrow \lambda = 0$. Substituting $\lambda=0$ back into the family equation, we get $x^2+y^2-8x-2y+7=0$. This is the equation of the first circle, which means the point $(1,2)$ actually lies on the first circle itself.

Let $(h,k)$ be an external point from which tangents are drawn to the circle $x^2+y^2=a^2$. The center of this circle is $O=(0,0)$ and its radius is $a$. Let the angle between the tangents be $\alpha = 60^\circ$. The angle between the line segment from the external point to the center ($d = \sqrt{h^2+k^2}$) and one of the tangents is $\alpha/2 = 30^\circ$. In the right-angled triangle formed by the center, the external point, and the point of tangency, we have $\sin(\alpha/2) = \frac{\text{radius}}{\text{distance from center to external point}}$. So, $\sin(30^\circ) = \frac{a}{\sqrt{h^2+k^2}}$. Since $\sin(30^\circ) = 1/2$, we have $1/2 = \frac{a}{\sqrt{h^2+k^2}}$. Squaring both sides and cross-multiplying, we get $\sqrt{h^2+k^2} = 2a \Rightarrow h^2+k^2 = (2a)^2 = 4a^2$. Replacing $(h,k)$ with $(x,y)$, the locus is $x^2+y^2=4a^2$. This is the director circle for tangents subtending a $60^\circ$ angle.

The general equation of a circle is $x^2+y^2+2Gx+2Fy+C=0$. For $S_1$, $G_1=-2, F_1=-3, C_1=5$. For $S_2$, $G_2=g, F_2=f, C_2=c$. The radical axis of $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$. $(x^2+y^2-4x-6y+5) - (x^2+y^2+2gx+2fy+c) = 0$ $(-4-2g)x + (-6-2f)y + (5-c) = 0$. This equation represents the line $x+y-1=0$. For these two equations to represent the same line, their coefficients must be proportional. Let the proportionality constant be $k$: $-4-2g = k \cdot 1 \quad \ldots (1)$ $-6-2f = k \cdot 1 \quad \ldots (2)$ $5-c = k \cdot (-1) \quad \ldots (3)$ From (1) and (2): $-4-2g = -6-2f \Rightarrow 2f-2g = -2 \Rightarrow f-g=-1 \quad \ldots (A)$ From (1) and (3): $-4-2g = -(5-c) = c-5 \Rightarrow c = 1-2g \quad \ldots (B)$ Circle $S_2$ passes through the point $(1,1)$. Substituting $x=1, y=1$ into the equation of $S_2$: $1^2+1^2+2g(1)+2f(1)+c=0 \Rightarrow 2+2g+2f+c=0 \quad \ldots (C)$ Now, substitute equations (A) and (B) into (C): $2+2g+2(g-1)+(1-2g)=0$ $2+2g+2g-2+1-2g=0$ $2g+1=0 \Rightarrow g = -1/2$. Using $g=-1/2$ in (A): $f = g-1 = -1/2-1 = -3/2$. Using $g=-1/2$ in (B): $c = 1-2(-1/2) = 1+1 = 2$. Finally, we need to find $g+f+c = (-1/2) + (-3/2) + 2 = -4/2 + 2 = -2+2 = 0$.

For the first circle $S_1: x^2+y^2=25$: its center is $C_1=(0,0)$ and its radius is $r_1=5$. For the second circle $S_2: x^2+y^2-8x-6y+21=0$: its center is $C_2=(-(-8)/2, -(-6)/2)=(4,3)$. Its radius is $r_2 = \sqrt{4^2+3^2-21} = \sqrt{16+9-21} = \sqrt{25-21} = \sqrt{4}=2$. The distance between the centers $d = C_1C_2 = \sqrt{(4-0)^2+(3-0)^2} = \sqrt{16+9} = \sqrt{25}=5$. To check if the circles intersect, we compare $d$ with $r_1+r_2$ and $|r_1-r_2|$. $r_1+r_2 = 5+2=7$. $|r_1-r_2| = |5-2|=3$. Since $|r_1-r_2| < d < r_1+r_2$ (i.e., $3 < 5 < 7$), the circles intersect. The angle $\theta$ between two intersecting circles is given by the formula $\cos\theta = \frac{r_1^2+r_2^2-d^2}{2r_1r_2}$. Substituting the values: $\cos\theta = \frac{5^2+2^2-5^2}{2(5)(2)} = \frac{25+4-25}{20} = \frac{4}{20} = \frac{1}{5}$. Therefore, the angle between the circles is $\theta = \cos^{-1}\left(\frac{1}{5}\right)$.

Let the centre be $(h,k)$. Since $k=2h$ and the circle passes through $(0,0)$ and $(8,6)$, we equate squared distances from centre: $h^2+k^2 = (h-8)^2+(k-6)^2$. This simplifies to $16h+12k=100$, or $4h+3k=25$. Substituting $k=2h$, we get $4h+6h=25 \implies 10h=25 \implies h=5/2$. Hence $k=5$. Centre is $(5/2,5)$. Radius squared $r^2 = (5/2)^2+5^2 = 25/4+25 = 125/4$. The equation is $(x-5/2)^2+(y-5)^2 = 125/4$, which simplifies to $x^2+y^2-5x-10y=0$.

The general tangent equation for $x^2+y^2+2gx+2fy+c=0$ at $(x_1, y_1)$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. For the circle $x^2+y^2-4x+6y-12=0$, we have $g=-2, f=3, c=-12$. At point $(5,1)$: $x(5)+y(1)+(-2)(x+5)+3(y+1)+(-12)=0$. This simplifies to $5x+y-2x-10+3y+3-12=0$. Combining terms gives $3x+4y-19=0$. This is the equation of the tangent.

The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1 - S_2 = 0$. Given $S_1: x^2+y^2+2x+3y-7=0$ and $S_2: x^2+y^2-2x-y+1=0$. Subtracting $S_2$ from $S_1$: $(x^2+y^2+2x+3y-7) - (x^2+y^2-2x-y+1) = 0$. This simplifies to $x^2+y^2+2x+3y-7 - x^2-y^2+2x+y-1 = 0$. Combining terms results in $(2x+2x) + (3y+y) + (-7-1) = 0$, which is $4x+4y-8=0$. Dividing by 4, the radical axis is $x+y-2=0$.

The circle is $x^2+y^2=25$, so its centre is $(0,0)$ and radius $r=5$. The external point is $P=(7,1)$. First, calculate the length of the tangent $PT$ from $P$ to the circle. $PT^2 = x_1^2+y_1^2-r^2 = 7^2+1^2-25 = 49+1-25 = 25$. So, $PT = 5$. The length of the chord of contact $L$ is given by the formula $L = \frac{2r \cdot PT}{\sqrt{r^2+PT^2}}$. Substitute $r=5$ and $PT=5$: $L = \frac{2 \times 5 \times 5}{\sqrt{5^2+5^2}} = \frac{50}{\sqrt{25+25}} = \frac{50}{\sqrt{50}}$. Simplify $\sqrt{50} = 5\sqrt{2}$. $L = \frac{50}{5\sqrt{2}} = \frac{10}{\sqrt{2}}$. Rationalizing the denominator: $L = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$. Thus, the length of the chord of contact is $5\sqrt{2}$.

Let the centre of the circle be $(h,k)$. Since it lies on the $x$-axis, $k=0$, so the centre is $(h,0)$. The circle passes through $(3,4)$ and $(1,0)$. The distance from the centre to both points must be equal to the radius, $r$. Thus, $(3-h)^2 + (4-0)^2 = r^2$ and $(1-h)^2 + (0-0)^2 = r^2$. Equating these: $(3-h)^2 + 16 = (1-h)^2$. This expands to $9 - 6h + h^2 + 16 = 1 - 2h + h^2$. Simplifying yields $25 - 6h = 1 - 2h$, which means $24 = 4h \implies h = 6$. The centre is $(6,0)$. Now, calculate the radius from $(1,0)$ to $(6,0)$: $r = \sqrt{(1-6)^2 + (0-0)^2} = \sqrt{(-5)^2} = \sqrt{25} = 5$. The radius of the circle is $5$.

Let the center of the circle be $(h,k)$. Since the circle passes through $(0,0)$ and $(0,6)$, the perpendicular bisector of the chord connecting these points must pass through the center. The midpoint of the chord is $(0,3)$. The chord is vertical, so its perpendicular bisector is the horizontal line $y=3$. Thus, $k=3$. Since the center $(h,3)$ lies on the line $2x+3y=15$, substitute $k=3$: $2h+3(3)=15 \Rightarrow 2h+9=15 \Rightarrow 2h=6 \Rightarrow h=3$. So, the center of the circle is $(3,3)$. The radius squared of the circle is $r^2 = (3-0)^2 + (3-0)^2 = 9+9 = 18$. The equation of the circle is $(x-3)^2+(y-3)^2=18$. To find the equation of the tangent at $(0,0)$, we can use the formula $T=0$: $(x-h)(x_1-h) + (y-k)(y_1-k) = r^2$. Substituting $(x_1,y_1)=(0,0)$, $(h,k)=(3,3)$, and $r^2=18$: $(x-3)(0-3) + (y-3)(0-3) = 18 \Rightarrow -3(x-3) -3(y-3) = 18 \Rightarrow -3x+9-3y+9 = 18 \Rightarrow -3x-3y+18=18 \Rightarrow -3x-3y=0$. Dividing by $-3$, we get $x+y=0$. Thus, option C is correct.

The equation of the first circle is $S_1: x^2 + y^2 - 16x - 20y + 100 = 0$. Its center $C_1 = (-(-16)/2, -(-20)/2) = (8, 10)$ and radius $r_1 = \sqrt{8^2 + 10^2 - 100} = \sqrt{64 + 100 - 100} = \sqrt{64} = 8$. The equation of the second circle is $S_2: (x-4)^2 + (y-7)^2 = 36$. Expanding this gives $x^2 - 8x + 16 + y^2 - 14y + 49 = 36 \Rightarrow x^2 + y^2 - 8x - 14y + 29 = 0$. Its center $C_2 = (4, 7)$ and radius $r_2 = \sqrt{36} = 6$. First, check the condition for intersection: the distance between centers $C_1C_2 = \sqrt{(8-4)^2 + (10-7)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$. Since $|r_1-r_2| = |8-6| = 2$ and $r_1+r_2 = 8+6 = 14$, and $2 < 5 < 14$, the circles intersect at two distinct points. The equation of the common chord of two intersecting circles $S_1=0$ and $S_2=0$ is given by $S_1 - S_2 = 0$. $(x^2 + y^2 - 16x - 20y + 100) - (x^2 + y^2 - 8x - 14y + 29) = 0$. Simplifying the equation: $(-16x - (-8x)) + (-20y - (-14y)) + (100 - 29) = 0$ $-16x + 8x - 20y + 14y + 71 = 0$ $-8x - 6y + 71 = 0$ Multiplying by $-1$ to get positive coefficients for $x$ and $y$: $8x + 6y - 71 = 0$. Thus, option C is correct.

Let the first circle be $C_1: x^2 + y^2 = 25$ with center $O=(0,0)$ and radius $R=5$. Let the second circle be $C_2: (x-1)^2 + (y-2)^2 = 1$ with center $P=(1,2)$ and radius $r=1$. Let $M(h,k)$ be the midpoint of a chord of $C_1$. The equation of the chord of $C_1$ whose midpoint is $(h,k)$ is given by $T=S_1'$, where $S_1'$ is the value of $S_1$ at $(h,k)$. So, $xh + yk - 25 = h^2 + k^2 - 25$, which simplifies to $xh + yk = h^2 + k^2$. This chord touches the circle $C_2$. This means the perpendicular distance from the center $P(1,2)$ of $C_2$ to the chord must be equal to its radius $r=1$. The equation of the chord can be written as $hx+ky-(h^2+k^2)=0$. Using the formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax+By+C=0$: $d = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$. Here, $(x_0, y_0) = (1,2)$ and the line is $hx+ky-(h^2+k^2)=0$. So, the distance is: $\frac{|h(1) + k(2) - (h^2+k^2)|}{\sqrt{h^2+k^2}} = 1$. To eliminate the absolute value and square root, square both sides of the equation: $(h + 2k - (h^2+k^2))^2 = (h^2+k^2)$. To find the locus, replace $(h,k)$ with $(x,y)$: $(x + 2y - (x^2+y^2))^2 = x^2+y^2$. This can be rearranged as $((x^2+y^2) - (x+2y))^2 = x^2+y^2$. This precisely matches option A.

Let the general equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle passes through the origin $(0,0)$, substituting these coordinates gives $0^2 + 0^2 + 2g(0) + 2f(0) + c = 0$, which implies $c=0$. Since the circle makes an intercept of length $a$ on the positive x-axis, it passes through $(a,0)$. Substituting $(a,0)$ into the circle's equation ($c=0$): $a^2 + 0^2 + 2g(a) + 2f(0) + 0 = 0 \Rightarrow a^2 + 2ga = 0$. Since $a \neq 0$, we get $a + 2g = 0 \Rightarrow 2g = -a$. Similarly, it makes an intercept of length $b$ on the positive y-axis, so it passes through $(0,b)$. Substituting $(0,b)$ into the circle's equation ($c=0$): $0^2 + b^2 + 2g(0) + 2f(b) + 0 = 0 \Rightarrow b^2 + 2fb = 0$. Since $b \neq 0$, we get $b + 2f = 0 \Rightarrow 2f = -b$. Substituting the values of $2g$, $2f$, and $c$ back into the general equation, we get $x^2 + y^2 - ax - by = 0$.

Option A, C, and D do not satisfy the conditions derived (e.g., passing through the origin or correctly matching the intercepts).

Let the two circles be $S_1: x^2 + y^2 + 2x + 3y + 1 = 0$ and $S_2: x^2 + y^2 + 4x + 3y + 2 = 0$. The equation of the common chord (radical axis) is $S_1 - S_2 = 0$. $(x^2 + y^2 + 2x + 3y + 1) - (x^2 + y^2 + 4x + 3y + 2) = 0$ $-2x - 1 = 0 \Rightarrow 2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$.

For circle $S_1$, the center $C_1 = (-1, -\frac{3}{2})$ and radius $r_1 = \sqrt{(-1)^2 + (-\frac{3}{2})^2 - 1} = \sqrt{1 + \frac{9}{4} - 1} = \sqrt{\frac{9}{4}} = \frac{3}{2}$. The distance $d$ from the center $C_1$ to the common chord $x = -\frac{1}{2}$ is $d = |-1 - (-\frac{1}{2})| = |-1 + \frac{1}{2}| = |-\frac{1}{2}| = \frac{1}{2}$.

The length of half the common chord, say $L/2$, can be found using the Pythagorean theorem: $(L/2)^2 = r_1^2 - d^2$. $L/2 = \sqrt{r_1^2 - d^2} = \sqrt{(\frac{3}{2})^2 - (\frac{1}{2})^2} = \sqrt{\frac{9}{4} - \frac{1}{4}} = \sqrt{\frac{8}{4}} = \sqrt{2}$. The total length of the common chord is $2 \times (L/2) = 2\sqrt{2}$.

Options B, C, and D provide incorrect lengths, as they do not result from the correct application of the distance and Pythagorean formulas.

Let the two circles be $S_1: x^2 + y^2 + 2gx + 2fy + c = 0$ and $S_2: x^2 + y^2 + 2g'x + 2f'y + c' = 0$. Their centers are $C_1(-g, -f)$ and $C_2(-g', -f')$. Their radii are $r_1 = \sqrt{g^2+f^2-c}$ and $r_2 = \sqrt{g'^2+f'^2-c'}$, respectively. If two circles cut each other orthogonally, their tangents at the point of intersection are perpendicular. This implies that the radii drawn to the point of intersection are perpendicular. Therefore, the triangle formed by the two centers and a point of intersection (P) is a right-angled triangle with the right angle at P. By the Pythagorean theorem, the square of the distance between the centers equals the sum of the squares of their radii: $(C_1C_2)^2 = r_1^2 + r_2^2$. The distance squared between centers $C_1$ and $C_2$ is: $(C_1C_2)^2 = (-g - (-g'))^2 + (-f - (-f'))^2 = (g' - g)^2 + (f' - f)^2$ $= g'^2 - 2gg' + g^2 + f'^2 - 2ff' + f^2$. The sum of the squares of the radii is: $r_1^2 + r_2^2 = (g^2+f^2-c) + (g'^2+f'^2-c')$. Equating these two expressions: $g'^2 - 2gg' + g^2 + f'^2 - 2ff' + f^2 = g^2+f^2-c + g'^2+f'^2-c'$. Cancelling terms $g^2, f^2, g'^2, f'^2$ from both sides: $-2gg' - 2ff' = -c - c'$. Multiplying by $-1$, we get the condition for orthogonality: $2gg' + 2ff' = c + c'$.