Circles & Lines

Why it works. The vector $(A, B)$ is perpendicular to the line: moving along the line by $(B, -A)$ leaves $Ax + By$ unchanged. So the expression $Ax + By + C$ is $0$ on the line and changes only as you move ACROSS it, by exactly $\sqrt{A^2+B^2}$ for every unit you move perpendicular to the line. Dividing its value at $(x_0, y_0)$ by that rate gives the perpendicular distance. Two parallel lines $Ax + By = C_1$ and $Ax + By = C_2$ are the level sets $C_1$ and $C_2$ of the same expression, so they sit $\tfrac{|C_1 - C_2|}{\sqrt{A^2+B^2}}$ apart — provided both are written with the SAME $A$ and $B$.

Spotting it. The coefficient pairs $(3, -4)$ and $(6, -8)$ are proportional, so the lines are parallel. The parallel-line formula needs the SAME $A$ and $B$ in both equations, so match them first.

Step 1. Divide the second equation by $2$: $3x - 4y = -\tfrac12$.

Step 2. Now $C_1 = 7$, $C_2 = -\tfrac12$ and $\sqrt{A^2 + B^2} = 5$, so the gap is $\dfrac{|7 - (-\tfrac12)|}{5} = \dfrac{15/2}{5} = \dfrac32$.

Check. The point $(1, -1)$ is on the first line, since $3 + 4 = 7$. Its distance to $6x - 8y + 1 = 0$ is $\dfrac{|6 + 8 + 1|}{10} = \dfrac{15}{10} = \dfrac32$ ✓

Spotting it. "Tangent" is a distance condition: the centre's distance to the line equals the radius. You could substitute and set a discriminant to zero, but the distance formula is one line.

Step 1. Write the line as $mx - y = 0$. Its distance from the centre $(5, 0)$ is $\dfrac{|5m|}{\sqrt{m^2 + 1}}$.

Step 2. Set this equal to $3$ and square: $25m^2 = 9m^2 + 9$, so $m^2 = \tfrac{9}{16}$ and $m = \pm\tfrac34$.

Check. The tangent from the origin has length $\sqrt{5^2 - 3^2} = 4$, so the right triangle formed by the origin, the centre and a point of tangency has legs $4$ (along the tangent) and $3$ (the radius). The tangent therefore makes an angle whose tangent is $\tfrac34$ with the line to the centre, which is the $x$-axis ✓

Spotting it. Tangent to both axes in the first quadrant puts the centre at $(r, r)$, so there is one unknown. The third tangency is a distance equation.

Step 1. The distance from $(r, r)$ to $5x + 12y - 60 = 0$ is $\dfrac{|5r + 12r - 60|}{13} = \dfrac{|17r - 60|}{13}$.

Step 2. Set it equal to $r$: $|17r - 60| = 13r$. The absolute value splits the problem in two. If the centre is on the origin's side of the line, $60 - 17r = 13r$ and $r = 2$. If it is on the far side, $17r - 60 = 13r$ and $r = 15$.

Step 3. Both circles exist. The radius-$2$ circle is the incircle of the $5$-$12$-$13$ triangle the line cuts off, which checks against the right-triangle inradius $\tfrac{a + b - c}{2} = \tfrac{5 + 12 - 13}{2} = 2$ ✓. The radius-$15$ circle sits beyond the hypotenuse. The absolute value is where the second answer lives, so read the question for which one it wants.

Why it works. A point lies on a bisector exactly when it is equally far from the two lines, and the distance formula turns that into one equation. Dropping the absolute values leaves a sign to choose. The expression $A x + By + C$ is positive on one side of its line and negative on the other, so the "$+$" equation collects the points where the two signed expressions agree and the "$-$" equation the points where they disagree: one bisector each. To tell them apart, test a point you know lies inside the angle you want.

Spotting it. Both lines pass through the origin, so both bisectors do too. The only real work is choosing the sign.

Step 1. Normalise: the signed distances are $\dfrac{4x - 3y}{5}$ and $\dfrac{5x - 12y}{13}$.

Step 2. Test $(1, 1)$: $4 - 3 = 1 > 0$ but $5 - 12 = -7 < 0$. The signed expressions DISAGREE inside this angle, so take the minus sign: $\dfrac{4x - 3y}{5} = -\dfrac{5x - 12y}{13}$.

Step 3. Clear denominators: $52x - 39y = -25x + 60y$, so $77x = 99y$ and the bisector is $y = \tfrac79x$.

Check by vectors. Unit vectors along the two lines are $\left(\tfrac35, \tfrac45\right)$ and $\left(\tfrac{12}{13}, \tfrac{5}{13}\right)$. Adding two unit vectors gives a rhombus, whose diagonal bisects the angle; the sum is $\left(\tfrac{99}{65}, \tfrac{77}{65}\right)$, and $\tfrac{77}{99} = \tfrac79$ ✓

Spotting it. $OB = AB = 5$, so the triangle is isosceles and the line $x = 3$ is a line of symmetry. That is one bisector for free (the symmetry shortcut); a second finishes the job.

Step 1. The bisector at $O$ lies between the $x$-axis and $4x - 3y = 0$. Inside the triangle both $4x - 3y$ and $y$ are positive, so equal distances read $\dfrac{4x - 3y}{5} = y$, giving $4x = 8y$, i.e. $y = \tfrac x2$.

Step 2. Meet it with $x = 3$: the incentre is $I = \left(3, \tfrac32\right)$, and the inradius is its distance to the $x$-axis, $\tfrac32$.

Check with $r = \dfrac{\text{Area}}{s}$: the area is $\tfrac12 \cdot 6 \cdot 4 = 12$ and $s = \tfrac{5 + 5 + 6}{2} = 8$, so $r = \tfrac{12}{8} = \tfrac32$ ✓

Why it works. The perpendicular from the centre to a chord bisects the chord, because the two radii to its ends make an isosceles triangle. A radius meets the tangent at its point of contact at a right angle. For two circles, the radii to a common tangent are both perpendicular to it, hence parallel; slide the tangent segment until one end sits at a centre, and it becomes a leg of a right triangle whose hypotenuse is the centre distance $d$ and whose other leg is the difference of the radii (external tangent) or their sum (internal tangent). Every formula in the box is the Pythagorean theorem in one of these triangles.

Spotting it. A chord length with a known centre: find the centre's distance to the line, then use half-chord $= \sqrt{r^2 - d^2}$. Solving for the two intersection points would work but drags radicals through a quadratic.

Step 1. $d = \dfrac{|1 + 2 - 7|}{\sqrt2} = \dfrac{4}{\sqrt2} = 2\sqrt2$, so $d^2 = 8$.

Step 2. Half the chord is $\sqrt{25 - 8} = \sqrt{17}$, so the chord is $2\sqrt{17}$.

Spotting it. Draw the radii to the points of tangency: both are perpendicular to the tangent, so they are parallel. Slide the tangent segment by the smaller radius so that it starts at a centre.

External. Both circles lie on the same side of the tangent, so the radii point the same way. After sliding, the leg along the larger radius is $8 - 3 = 5$, and the tangent is $\sqrt{13^2 - 5^2} = 12$.

Internal. The tangent passes between the circles, so the radii point opposite ways and the leg becomes $8 + 3 = 11$: the length is $\sqrt{169 - 121} = \sqrt{48} = 4\sqrt3$.

Spotting it. The centre distance and the radius together give the power of $P$: for EVERY line through $P$, $PA \cdot PB = d^2 - r^2$, which is also the square of the tangent length.

Step 1. The power is $13^2 - 5^2 = 144$ (so a tangent from $P$ has length $12$).

Step 2. $PA \cdot PB = 144$ with $PA = 9$ gives $PB = 16$.

Step 3. $A$ and $B$ lie on the same side of $P$, so $AB = PB - PA = 7$.

Check that such a line exists. The chord's midpoint is $\tfrac{9 + 16}{2} = 12.5$ from $P$, less than $13$, so the centre is $\sqrt{13^2 - 12.5^2} = \sqrt{12.75}$ from the line, which is less than the radius $5$ ✓

Why it works. After substituting the parabola, $x$ appears only as $x^2$, so each value $u$ the equation produces must be split back into $x = \pm\sqrt u$, and the parabola then fixes $y$. A positive $u$ gives the mirror pair $(\pm\sqrt u, y)$, $u = 0$ gives one point on the axis, and a negative $u$ gives nothing. That is why the SIGNS of the roots, not just how many there are, decide the count.

Spotting it. Both curves are symmetric under $x \to -x$, so intersection points come in mirror pairs. An ODD count forces a point on the $y$-axis, and the only point of the parabola there is its vertex $(0, -2)$.

Step 1. For the vertex to lie on the circle, $r = 2$.

Step 2. Confirm the count. With $u = x^2$ and $y = u - 2$: $u + (u - 2)^2 = 4$, i.e. $u^2 - 3u = 0$. So $u = 0$ (one point, the vertex) or $u = 3$ (two points, $x = \pm\sqrt3$, $y = 1$): three points in all ✓

Spotting it. "Distance to the focus" should trigger the focus–directrix property, which trades a square-root distance for a vertical one.

Step 1. $x^2 = 8y$ is $x^2 = 4py$ with $p = 2$: focus $(0, 2)$, directrix $y = -2$.

Step 2. The distance to the focus equals the distance to the directrix, which is $y + 2$. So $y + 2 = 10$ and $y = 8$.

Step 3. $x^2 = 64$, so $P = (\pm8, 8)$. Check: from $(8, 8)$ to $(0, 2)$ is $\sqrt{64 + 36} = 10$ ✓

Spotting it. Four points means two DIFFERENT POSITIVE values of $u = x^2$. That is a question about the signs of a quadratic's roots, which Vieta and the discriminant answer without solving.

Step 1. Substitute: $u + (u + k)^2 = 9$, i.e. $u^2 + (2k + 1)u + (k^2 - 9) = 0$.

Step 2. Two distinct real roots: the discriminant $(2k + 1)^2 - 4(k^2 - 9) = 4k + 37$ must be positive, so $k > -\tfrac{37}{4}$.

Step 3. Both roots positive: the product $k^2 - 9$ and the sum $-(2k + 1)$ must both be positive, so $k < -3$.

Step 4. The integers with $-9.25 < k < -3$ are $-9, -8, \dots, -4$: six values. Check the ends. At $k = -3$ one root is $u = 0$ and the count drops to three; at $k = -9$ the roots are $u = 8$ and $u = 9$, giving four points ✓

Every conic question reduces to completing the square and reading off $a$, $b$, $c$. For $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$: $a = 5$, $b = 3$, $c = 4$, so the foci are $(\pm4, 0)$ and the eccentricity is $\tfrac45$. Sign discipline decides which axis is major: the LARGER denominator sits under the variable along the major axis, whichever letter that is.

These turn geometry into one equation. "The sum of distances from $P$ to $(\pm4,0)$ is $10$" IS the ellipse above, with no algebra. On a contest the focal definition usually beats expanding the equation, because it names $a$ and $c$ immediately and the rest follows from $b^2 = a^2 - c^2$.

Step 1 — $a^2 = 169$, $b^2 = 144$, so $a = 13$, $b = 12$.

Step 2 — $c^2 = a^2 - b^2 = 25$, so $c = 5$ and the foci are $(\pm5, 0)$.

Step 3 — eccentricity $e = \tfrac ca = \tfrac{5}{13}$. Check: $5^2 + 12^2 = 13^2$, the same right triangle that always relates $a$, $b$, $c$ for an ellipse.

Spotting it. A point and both foci are given, so the focal definition hands you $2a$ with no equation to expand.

Step 1. $PF_2 = \tfrac{16}{5}$ (straight up from $(3, 0)$), and $PF_1 = \sqrt{6^2 + \left(\tfrac{16}{5}\right)^2} = \sqrt{\tfrac{1156}{25}} = \tfrac{34}{5}$.

Step 2. $2a = \tfrac{16}{5} + \tfrac{34}{5} = 10$, so $a = 5$. With $c = 3$, $b = \sqrt{25 - 9} = 4$.

Step 3. The area is $\pi ab = 20\pi$. Check that $P$ is on $\tfrac{x^2}{25} + \tfrac{y^2}{16} = 1$: $\tfrac{9}{25} + \tfrac{256/25}{16} = \tfrac{9}{25} + \tfrac{16}{25} = 1$ ✓

Spotting it. For a hyperbola the focal DIFFERENCE is fixed: $|PF_1 - PF_2| = 2a = 6$.

Step 1. A point on the right branch is nearer the right focus, so $PF_1 - PF_2 = 6$ and $PF_2 = 14 - 6 = 8$.

Step 2. The other sign would give $PF_2 = 20$, which describes a point nearer the LEFT focus, on the left branch. Check: $PF_1^2 - PF_2^2 = (x + 5)^2 - (x - 5)^2 = 20x$, so $20x = 196 - 64 = 132$ and $x = 6.6$; then $y^2 = 16\left(\tfrac{6.6^2}{9} - 1\right) = 61.44$ and $PF_2^2 = 1.6^2 + 61.44 = 64$ ✓