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प्रश्न
The shortest distance from the point (2, –7) to the circle x2 + y2 – 14x – 10y – 151 = 0 is equal to 5.
पर्याय
True
False
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उत्तर
This statement is False.
Explanation:
Given equation of circle is x2 + y2 – 14x – 10y – 151 = 0
Shortest distance = distance between the point (2, – 7)
And the centre – radius of the circle
Centre of the given circle is
2g = – 14 ⇒ g = – 7
2f = – 10 ⇒ f = – 5
∴ Centre = (– g, – f) = (7, 5)
And r = `sqrt((-7)^2 + (-5)^2 + 151)`
= `sqrt(49 + 25 + 151)`
= `sqrt(225)`
= 15
∴ Shortest distance = `sqrt((7 - 2)^2 + (5 + 7)^2) - 15`
= `sqrt(25 + 144) - 15`
= 13 – 15
= |– 2|
= 2
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