Question

Find equation of circle tangent to 3x+y+3=0 at (-3, 6)& tangent to x+3y-7=0

Answer 1

Dear student

$$\begin{gathered} Thecircleis\tan gentto3x+y+3=0 \\ eqnofnormalthroughcentreis \\ x-3y=k \\ itpassesthrough(-3,6) \\ sox-3y=-21 \\ letthecentrebe(3k-21,k) \\ centreisequidis\tan tfrom\tan gents3x+y+3=0andx+3y-7=0 \\ so \\ \frac{\left|3(3k-21)+k+3\right|}{\sqrt{3^2+1^2}}=\frac{\left|(3k-21)+3k-7\right|}{\sqrt{1^2+3^2}} \\ \frac{\left|10k-60\right|}{\sqrt{10}}=\frac{\left|6k-28\right|}{\sqrt{10}} \\ \left|10k-60\right|=\left|6k-28\right| \\ 10k-60=\pm (6k-28) \\ solvingwegetk=8and11/2 \\ centreis(3,8)and(-9/2,11/2)radiusis2\sqrt{10}and\sqrt{10}/2 \\ circlesare \\ (x-3{)}^2+(y-8{)}^2=40 \\ (x+9/2{)}^2+(y-11/2{)}^2=10/4 \end{gathered}$$

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