座標幾何：直線と円

For two points $A(x_1,y_1)$ and $B(x_2,y_2)$, the gradient is the change in $y$ over the change in $x$: $m=\frac{y_2-y_1}{x_2-x_1}$. The midpoint of $AB$ is the average of the coordinates, $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$. The distance between the points comes from Pythagoras: $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$. The order of subtraction affects the sign of the gradient, but since the distance squares each difference, it gives the same value either way.

Given a gradient $m$ and one point $(x_1,y_1)$ on the line, its equation is $y-y_1=m(x-x_1)$ — the most flexible form to start from. Expanding and rearranging gives either $y=mx+c$ (with $y$-intercept $c$) or $ax+by+c=0$. When two points are given, first find the gradient, then substitute either point. Exam questions often ask for the answer in the form $ax+by+c=0$, so clearing fractions to obtain integer coefficients is a safe finishing move.

Two lines are parallel exactly when their gradients are equal, $m_1=m_2$. They are perpendicular exactly when the product of their gradients is $-1$, $m_1{m}_2=-1$. Equivalently, the gradient of a line perpendicular to a given one is the negative reciprocal: flip and change sign. For instance, a line perpendicular to one of gradient $\frac{2}{3}$ has gradient $-\frac{3}{2}$. The pair of a horizontal line (gradient $0$) and a vertical line (undefined gradient) is the special case this rule cannot capture — handle it by inspection.

A circle with centre $(a,b)$ and radius $r$ has equation $(x-a{)}^2+(y-b{)}^2=r^2$ — this is just the distance formula stating that every point is a fixed distance $r$ from the centre. Expanding gives $x^2+y^2-2ax-2by+(a^2+b^2-r^2)=0$, and comparing with the general form $x^2+y^2+2gx+2fy+c=0$ shows the centre is $(-g,-f)$ and the radius is $r=\sqrt{g^2+f^2-c}$. To recover the centre and radius from the general form, either apply that formula or complete the square separately in $x$ and in $y$.