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Question

The number of circles passes through (3,-6) and touches both the axes are


A
0
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B
1
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C
2
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D
4
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Solution

The correct option is B 0
Equation of circle touches both the axes is given by,
$$(x-a)^2+(y-a)^2 = a^2$$
Given it also passes through $$(3,-6)$$
$$\Rightarrow (3-a)^2+(a+6)^2=a^2\Rightarrow a^2+6a+45=0$$
Clearly this quadratic does not posses real roots.Hence no circle is possible.

Maths

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