﻿ If four distinct points (2, 3), (0, 2), (4, 5) and (0, t) are concylic, then t3+ 17 is equal to  : Kaysons Education

# If Four Distinct Points (2, 3), (0, 2), (4, 5) And (0, t) Are Concylic, Then t3+ 17 Is Equal To

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## Question

### Solution

Correct option is

4930

Let equation of the circle be x2 + y2 + 2gx + 2fy + c = 0. Since it passes through the points (2, 3), (0, 2), (4, 5) we have

4+ 6f + c = –13, 4f  + c = –4, 8g + 10f  + c = –41. Solving these equations we get g = 5/2, f = –19/2, c = 34 = 0. As the circle passes through the point (0, t), t2 – 19t + 34 = 0 ⇒ t = 2 or t = 17. But t = 2 corresponds to the point (0, 2) which different from (0, t). Therefore t = 17 and t3 + 17 = 4930.

#### SIMILAR QUESTIONS

Q1

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Equation of the chord is

Q2

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Q3

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Q4

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Q5

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Q6

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Q7

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Q8

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Q9

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Q10

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