Q.The circle $x^2+y^2=4x+8y+5$ intersects the line $3x-4y=m$ at two distinct points if

A circle is the locus of a point whose distance from a fixed centre is always the constant radius.

Standard form, centre $(h,k)$, radius $r$: $(x-h)^2+(y-k)^2=r^2$ (centre at the origin: $x^2+y^2=r^2$). Expanding gives the general form

with centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$. Any second-degree equation with equal, nonzero $x^2$/$y^2$ coefficients and no $xy$ term is always a circle of this kind. The value $g^2+f^2-c$ being $>0$, $=0$ or $<0$ gives a real circle, a single point, or no real locus at all.

Family of circles through a line–circle intersection (Theorem 5.1). For circle $S=0$ and line $L=0$, every circle through their intersection points is $S+\lambda L=0$ for some $\lambda\in\mathbb R$ — a single extra condition (e.g. "the chord $L$ is a diameter, so the centre lies on $L$") pins down $\lambda$.

Diameter form (Theorem 5.2). With diameter ends $(x_1,y_1),(x_2,y_2)$: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$ (from the semicircle right-angle property, $\angle APB=90^\circ$).

Position of a point (Theorem 5.3). Substituting $(x_1,y_1)$ into $x^2+y^2+2gx+2fy+c$: the point is outside/on/inside according as the value is $>0,=0,<0$.

Tangent and normal at $(x_1,y_1)$ on the circle:

The normal always passes through the centre — the single fact behind most "normal to a circle" problems.

Tangency of $y=mx+c$ to $x^2+y^2=a^2$: $c^2=a^2(1+m^2)$, tangent $y=mx\pm a\sqrt{1+m^2}$, point of contact $\left(\mp\frac{am}{\sqrt{1+m^2}},\pm\frac a{\sqrt{1+m^2}}\right)$.

Theorem 5.4. From any external point, exactly two tangents can be drawn (the slope-condition equation is a quadratic in $m$); from inside, both are imaginary; from on the circle, they coincide.