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  • The equation of circle passing through (3,-6) and touching both the axes isOption: 1 <img alt="\begin{aligned} & x^2+y^2-6 x+6 y+9=0 \\ \end

The equation of circle passing through (3,-6) and touching both the axes is

Option: 1

\begin{aligned} & x^2+y^2-6 x+6 y+9=0 \\ \end{aligned}


Option: 2

x^2+y^2+6 x-6 y+9=0 \\


Option: 3

x^2+y^2+30 x-30 y+225=0


Option: 4

x^2+y^2-30 x+30 y+225=0


Answers (1)

best_answer

Circle possible in IV quadrant. Equation of circle is 

(x-c)^2+(y+c)^2=c^2\: \: \: \: \: \: \: \: \: \: \: .....(i)

Its passes through (3,-6)

\Rightarrow \quad(3-c)^2+(c-6)^2=c^2

\Rightarrow \quad c^2-18 c+45=0

           (c-15)(c-3)=0

\therefore \: \: \: \: \: \: \: \: \mathrm{c=3,15}

From Eq. (i) , equation of circles are 

\begin{array}{r} x^2+y^2-6 x+6 y+9=0 \\ \text { and } \: \: x^2+y^2+30 x+30 y+225=0 \end{array}

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