The centre of a circle of radius $4 \sqrt{5}$ lies on the line y=x and satisfies the inequality 3x+6y>10. If the line x+2y=3 is a tangent to the circle, then the equation of the circle is

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Solution

The correct option is A
$c = (a, a) r a d i u s = 4 \sqrt{5} = l e n g t h o f t h e ⊥ f r o m (a, a) t o t h e l i n e i . e, \frac{| a + 2 (a) - 3 |}{\sqrt{4 + 1}} = \pm 4 \sqrt{5} \Rightarrow a = \frac{23}{3}, \frac{- 17}{3} ∴ C e n t r e (\frac{23}{3}, \frac{23}{3}) o r (\frac{- 17}{3}, \frac{- 17}{3}) 3 x + 6 y > 10 C = (\frac{23}{3}, \frac{23}{3}) ∴ {(x - \frac{23}{3})}^{2} + {(y - \frac{23}{3})}^{2} = 80$

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Similar questions

\frac{3}{2} x + \frac{2}{3} y = 13 \frac{1}{3}; \frac{2}{3} x + \frac{3}{2} y = 8 \frac{1}{3}

Equation of circle touching the line |x-2|+|y-3|=4 will be

x^{2} + y^{2} - 4 x + 6 y + 10 = 0

4 x - 3 y + 8 = 0

(x - 2)^{2} + (y - 3)^{2} = 32

\int_{- 2}^{6} [(x + 1) + \sqrt{32 - (x - 2)^{2}} + 3] d x

x + y = 6

x + 2 y = 4

(2, 6)

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