(a) At \(x=0\), \((y-5)^2=31\), so \(\boxed{y=5\pm\sqrt{31}}\).

(b) The midpoint of \(AB\) is \((0,5)\), and its half-length is \(\sqrt{31}\). Thus \(\boxed{x^2+(y-5)^2=31}\).

(a) Completing the squares gives \[(x-3)^2+(y+2)^2=40,\] so the centre is \(C(3,-2)\). The radius CP has gradient \((4+2)/(5-3)=3\), hence the tangent gradient is \(-1/3\). Its equation is \[y-4=-\frac13(x-5),\qquad y=-\frac{x}{3}+\frac{17}{3}.\] Thus \(A=(17,0)\), \(B=(0,17/3)\), and \[\boxed{\text{area of OAB}=\tfrac12(17)(17/3)=\frac{289}{6}}.\]

(b) The radius is \(r=\sqrt{40}\). Each side of the inscribed equilateral triangle is a chord subtending \(120^\circ\) at the centre, so \[\ell=2r\sin60^\circ=\sqrt{120}.\] Therefore \[\boxed{\text{area of PQR}=\frac{\sqrt3}{4}\ell^2=30\sqrt3}.\]

(a) \[m_{AB}=\frac{0-3}{3-(-2)}=-\frac35,\qquad m_{BC}=\frac{5-0}{6-3}=\frac53.\] Their product is \(-1\), so AB is perpendicular to BC and \(\angle ABC=90^\circ\).

(b) The right angle subtends diameter AC, so D is its midpoint: \[\boxed{D=(2,4)}.\]

(c) \(r^2=DA^2=(-2-2)^2+(3-4)^2=17\). Thus \[\boxed{(x-2)^2+(y-4)^2=17}.\]

(d) Since D is the midpoint of BE, \[E=2D-B=(4-3,8-0)=(1,8).\] The radius DE has gradient \((8-4)/(1-2)=-4\), so the tangent gradient is \(1/4\). Its equation is \[\boxed{y-8=\frac14(x-1)}\] or \(y=x/4+31/4\).

(a) Let the centre be C. \[CA^2=(5-1)^2+(11-5)^2=52,\] so A lies on the circle. The radius gradient is \((11-5)/(5-1)=3/2\), whose product with the gradient \(-2/3\) of l is \(-1\). Thus l is perpendicular to CA at A and is a tangent.

(b) The other centre lies the same distance from A on the opposite side: \[C_2=2A-C=(2-5,10-11)=(-3,-1).\] The required equation is \[\boxed{(x+3)^2+(y+1)^2=52}.\]

(a) Perpendicular vectors give \((p-6,3)\cdot(14-p,11)=0\), so \((p-6)(14-p)+33=0\). This yields \((p-3)(p-17)=0\). Since \(p<10\), \(\boxed{p=3}\).

(b) The right angle at \(B\) means \(AC\) is a diameter. Its midpoint is \((10,11)\), and \(r^2=(14-10)^2+(18-11)^2=65\). Thus \(\boxed{(x-10)^2+(y-11)^2=65}\).

(c) The radius to \(C\) is \((4,7)\), a normal to the tangent. Hence \(4(x-14)+7(y-18)=0\), or \(\boxed{4x+7y-182=0}\).

(a) On the x-axis, \(y=0\), so \[x^2-4x-77=(x+7)(x-11)=0.\] Hence \[\boxed{x=-7\quad\text{and}\quad x=11}.\]

(b) Completing squares gives \[(x-2)^2+(y+3)^2=90,\] so the centre is \(C(2,-3)\). Take \(A(-7,0)\), \(B(11,0)\). The radius gradients are \(-1/3\) at A and \(1/3\) at B, so the tangent gradients are 3 and \(-3\). Their equations are \[y=3(x+7)=3x+21,\qquad y=-3(x-11)=-3x+33.\] Equating gives \(6x=12\), so \[\boxed{\text{intersection}=(2,27)}.\]

(a) AB is vertical and BC horizontal, so \(\angle ABC=90^\circ\). Thus AC is a diameter and its midpoint is the centre: \[\left(\frac{7+1}{2},\frac{1+9}{2}\right)=(4,5).\] The radius satisfies \[r^2=(7-4)^2+(1-5)^2=25.\] Hence \[\boxed{(x-4)^2+(y-5)^2=25}.\]

(b) The radius to B has gradient \((9-5)/(7-4)=4/3\), so the tangent gradient is \(-3/4\). Therefore \[\boxed{y-9=-\frac34(x-7)}.\]

(a) The midpoint is \(M=(15/2,1/2)\). The gradient of AB is \((-1-2)/(10-5)=-3/5\), so the perpendicular gradient is \(5/3\). Thus \[y-\frac12=\frac53\left(x-\frac{15}{2}\right),\qquad y=\frac53x-12.\]

(b) \(AB^2=(10-5)^2+(-1-2)^2=34\). The circle is \[(x-5)^2+(y-2)^2=34.\]

The centre is \((8,10)\). The radius gradients at \((6,16)\) and \((14,8)\) are \((16-10)/(6-8)=-3\) and \((8-10)/(14-8)=-1/3\). The tangent gradients are therefore \(1/3\) and \(3\). Their equations are \[y-16=\frac13(x-6)\iff y=\frac x3+14,\qquad y-8=3(x-14)\iff y=3x-34.\] Thus \(P=(0,14)\), \(R=(34/3,0)\). Equating the tangent equations gives \(x/3+14=3x-34\), so \(Q=(18,20)\). Split the quadrilateral along \(OQ\): \[\text{area}=\text{area}(OPQ)+\text{area}(OQR)=\frac12(14)(18)+\frac12\left(\frac{34}{3}\right)(20)=\boxed{\frac{718}{3}}.\]

(a) Substitute the first tangent into the circle: \[(x-6)^2+(2x-2)^2=20\iff5(x-2)^2=0.\] Thus \(A=(2,5)\). For the second tangent, \[(x-6)^2+(-2x+22)^2=20\iff5(x-10)^2=0,\] giving \(\boxed{A=(2,5),\quad B=(10,5)}\).

(b) The centre is \(C=(6,3)\), radius \(\sqrt{20}\). The chord length is \(8\), and its perpendicular distance from \(C\) is \(2\). The minor central angle is \[\alpha=2\tan^{-1}(4/2)=2\tan^{-1}2=2.214297\ldots\text{ radians}.\] Subtract triangle \(ACB\) from the minor sector: \[\begin{aligned}A&=\frac12(20)\alpha-\frac12(8)(2)\\&=20\tan^{-1}2-8=14.142974\ldots\\&=\boxed{14.1}\quad(3\text{ s.f.}).\end{aligned}\]

(a) \((x-2)^2+(y-6)^2=100\).

(b) \[36+(k-6)^2=100\Rightarrow k-6=\pm8\Rightarrow k=14,-2.\]

(a) The centre is \((r+5,r+8)\).

(b) \[(r+5)^2+(r+8)^2=225\Rightarrow r^2+13r-68=(r+17)(r-4)=0.\] Since \(r>0\), \(r=4\), centre \((9,12)\).

(c) The furthest point lies on the ray from the origin through the centre, so its radius gradient is \(12/9=4/3\). The tangent is perpendicular: gradient \(-3/4\).

(a) Completing squares gives \(C_1:(x+3)^2+(y-5)^2=16\), centre (−3,5), radius 4. C₂ has centre (9,−4), radius 8. Centre distance \(\sqrt{12^2+(-9)^2}=15\). (b) The circles are disjoint. Greatest distance 15+4+8=27; least 15−4−8=3.