Question

Let "p' be a root of the equation x^2 - 5x+7=0, then
the area of circle with centre at (P, P)'and passing
through point (1, 4) is
1. 3π sq, units
2 . 5π sq. units
3 7π sq. units
4. None of these
​

Answer 1

answer : option (1) 3π sq. unit

centre of circle is (p, p) and it is passing through point (1,4).

radius of circle is ....

r = √{(p - 1)² + (p - 4)²}

= √{p² - 2p + 1 + p² - 8p + 16}

= √{2p² - 10p + 17}

= √{2(p² - 5p + 7) + 3}........(1)

as it is given, p is a root of x² - 5x + 7

so, p² - 5p + 7 = 0, ......(2)

putting equation (2) in equation (1),

r = √{2 × 0 + 3} = √3 unit

now area of circle, A = πr² = π(√3)² = 3π sq unit.

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Answer 2

Answer:

4. None of these

Step-by-step explanation:

The equation $x^2 - 5x + 7 =0$ has no real roots since $D < 0$.

Hence, p must be imaginary.

Therefore, such a circle does not exist.

Therefore, the answer is 4. None of These