Concept understanding — Circle — Standard Equation and Properties
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=r2 (centre at the origin: x2+y2=r2). Expanding gives the general form
x2+y2+2gx+2fy+c=0,
with centre (−g,−f) and radius g2+f2−c. Any second-degree equation with equal, nonzero x2/y2 coefficients and no xy term is always a circle of this kind. The value g2+f2−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+λL=0 for some λ∈R — a single extra condition (e.g. "the chord L is a diameter, so the centre lies on L") pins down λ.
Diameter form (Theorem 5.2). With diameter ends (x1,y1),(x2,y2): (x−x1)(x−x2)+(y−y1)(y−y2)=0 (from the semicircle right-angle property, ∠APB=90∘).
Position of a point (Theorem 5.3). Substituting (x1,y1) into x2+y2+2gx+2fy+c: the point is outside/on/inside according as the value is >0,=0,<0.
Force the two circle conditions — coefficient of x2= coefficient of y2, and coefficient of xy=0 — onto the given equation to pin down p,q, then substitute back to get the actual circle.
Step 1. Coefficient of xy must vanish.(3−p)=0⇒p=3.
Step 2. Coefficient of x2 must equal coefficient of y2.3=q⇒q=3.
Step 3. Substitute p=3,q=3 into the original equation.
3x2+(3−3)xy+3y2−2(3)x=8(3)(3)⇒3x2+3y2−6x=72.
Step 4. Divide by 3 and rearrange to general form. …