1.
Tâm I (2,1)
Bán kính R = 5
2.
Tt của (I) tại A (5,3)
=> Tt có vtpt $\vec{IA}=(3,2)$
Tt (C):$\left \{ {{qua A(5,3)} \atop {vtpt\vec{CA}=(3,2)}} \right.$
=> pttq là:
$3(x-5)+2(y-3)=0$
<=>$3x+2y-21=0$
3. Do tt // d1
=> tt có dạng: $5x-2y+m=0$(m#2)
Ta có $(d) (I,(tt)) = R$
<=>$\dfrac{|5x_I-12y_I+m|}{\sqrt{5^2+(-12)^2}}=5$
<=>$\dfrac{|5.2-12.1+m|}{13}=5$
<=>$|-2+m|=65$
<=> \(\left[ \begin{array}{l}-2+m=65\\-2+m=-65\end{array} \right.\)
<=>\(\left[ \begin{array}{l}m=67(n)\\m=-63(n)\end{array} \right.\)
=> (tt1): $5x-12y+67=0$
(tt2): $5x-12y-63=0$
4. Do $d\perp d_2$
=> tt có dạng $4x-3y+m=0$
Ta có: $(d) (I, (tt)) = R$
<=>$\dfrac{|4x_I-3y_I-m|}{\sqrt{4^2+(-3)^2}}=5$
<=>$\dfrac{|4.2-3.1-m|}{5}=5$
<=>$|5-m|=25$
<=>\(\left[ \begin{array}{l}5-m=25\\5-m=-25\end{array} \right.\)
<=>\(\left[ \begin{array}{l}m=-20(n)\\m=30(n)\end{array} \right.\)
=> (tt1): $12x+5y-20=0$
(tt2): $12x+5y+30=0$
