Question

Obtenha a equação reduzida da circunferência H que passa pelos pontos (4, 2),
(6, 22) e (22, 26).

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm equac_{\!\!,}\tilde ao\,da\,reta\,que\,passa}\\\underline{\rm\,por\,A(4,2)~e~B(6,22):}\\\sf m=\dfrac{22-2}{6-4}=\dfrac{20}{2}=10.\\\sf y=2+10(x-4)\\\sf y=2+10x-40\\\sf y=10x-38\implies 10x-y=38\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Equac_{\!\!,}\tilde ao\,da\,reta\,que\,passa\,por}\\\underline{\rm B(6,22)~e~C(22,26):}\\\sf m=\dfrac{26-22}{22-16}=\dfrac{4\div2}{6\div2}=\dfrac{2}{3}\\\\\sf y=22+\dfrac{2}{3}(x-6)\\\\\sf y=22+\dfrac{2}{3}x-4\cdot3\\\\\sf 3y=66+2x-12\\\sf2x-3y=-54\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm intersecc_{\!\!,}\tilde ao\,entre\,as\,retas:}\\\begin{cases}\sf10x-y=38\cdot(-3)\\\sf2x-3y=-54\end{cases}\\\\\begin{cases}\sf -30x+3y=-114\\\sf 2x-3y=-54\end{cases}\\\sf -28x=-168\cdot(-1)\\\sf 28x=168\\\sf x=\dfrac{168}{28}\\\sf x=6\\\sf 10x-y=38\\\sf y=10x-38\\\sf y=10\cdot6-38\\\sf y=60-38\\\sf y=22\end{array}}$

$\Large\boxed{\begin{array}{l}\sf C(6,22)\\\sf d_{C,A}=\sqrt{(22-4)^2+(26-2)^2}\\\sf d_{C.A}=\sqrt{18^2+24^2}\\\sf d_{C,A}=\sqrt{324+576}\\\sf d_{C,A}=\sqrt{900}\\\sf d_{C,A}=30\\\sf R=d_{C,A}\\\sf R=30\end{array}}$

$\Large\boxed{\begin{array}{l}\sf\underline{\rm Equac_{\!\!,}\tilde ao\,da\,circunfer\hat encia}\\\underline{\rm que\,passa\,por\,A(4,2),B(6,22)\,e\,C(22,26)\!:}\\\sf (x-6)^2+(y-22)^2=30^2\\\\\Large\boxed{\boxed{\boxed{\boxed{\sf (x-6)^2+(y-22)^2=900}}}}\end{array}}$