If the radius of the circle $x^{2} + y^{2} + a x + (1 - a) y + 5 = 0$
does not exceed 5, write the number of integral values a.

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Solution

Equation of given circle is
$x^{2} + y^{2} + a x + (1 - a) y + 5 = 0 g = \frac{a}{2}, f = \frac{(1 - a)}{2} R a d i u s = r = \sqrt{g^{2} + f^{2} - c} = \sqrt{\frac{a^{2}}{4} + \frac{(1 - a)^{2}}{4}} - 5 < 5$
[Radius can be at most 5]
$\frac{a^{2}}{4} + \frac{(1 - a)^{2}}{4} - 5 < 25 a^{2} + (1 - a)^{2} < 120$
Sum of two square should be less than 120 a can be at most 8 and atleast -7 So a can take 16 integral values which are from -7 to 8.

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