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प्रश्न
The equation of the circle passing through the origin which cuts off intercept of length 6 and 8 from the axes is
पर्याय
x2 + y2 − 12x − 16y = 0
x2 + y2 + 12x + 16y = 0
x2 + y2 + 6x + 8y = 0
x2 + y2 − 6x − 8y = 0
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उत्तर
x2 + y2 − 6x − 8y = 0
The centre of the required circle is \[\left( \frac{6}{2}, \frac{8}{2} \right) = \left( 3, 4 \right)\] .
The radius of the required circle is
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