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If x^2+y^2+px+y(1-p)=0 is the equation o...

If `x^2+y^2+px+y(1-p)=0` is the equation of circle `r in (0,5] , q=p^2` then number of integral value of (p,q) satisfy is

A

16

B

14

C

19

D

21

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the given equation of a circle and find the integral values of \( p \) and \( q \) that satisfy the conditions. ### Step-by-Step Solution: 1. **Given Equation**: The equation provided is \[ x^2 + y^2 + px + y(1 - p) = 0. \] We can rearrange this as: \[ x^2 + y^2 + px + (1 - p)y = 0. \] 2. **Identifying Circle Parameters**: The general form of a circle's equation is \[ x^2 + y^2 + 2gx + 2fy + c = 0, \] where \( g \) and \( f \) are related to the center of the circle, and \( c \) is a constant. By comparing coefficients, we identify: - \( 2g = p \) → \( g = \frac{p}{2} \) - \( 2f = 1 - p \) → \( f = \frac{1 - p}{2} \) - \( c = 0 \) 3. **Radius of the Circle**: The radius \( R \) of the circle can be calculated using the formula: \[ R = \sqrt{g^2 + f^2 - c}. \] Substituting the values we found: \[ R = \sqrt{\left(\frac{p}{2}\right)^2 + \left(\frac{1 - p}{2}\right)^2}. \] 4. **Simplifying the Radius**: \[ R = \sqrt{\frac{p^2}{4} + \frac{(1 - p)^2}{4}} = \sqrt{\frac{p^2 + (1 - 2p + p^2)}{4}} = \sqrt{\frac{2p^2 - 2p + 1}{4}} = \frac{\sqrt{2p^2 - 2p + 1}}{2}. \] 5. **Setting Up the Radius Condition**: We know that the radius must satisfy: \[ 0 < R \leq 5. \] Squaring both sides gives: \[ 0 < \frac{2p^2 - 2p + 1}{4} \leq 25. \] Multiplying through by 4 results in: \[ 0 < 2p^2 - 2p + 1 \leq 100. \] 6. **Solving the Inequalities**: - From \( 2p^2 - 2p + 1 > 0 \): This quadratic has no real roots (discriminant \( D = (-2)^2 - 4 \cdot 2 \cdot 1 < 0 \)), so it is always positive. - From \( 2p^2 - 2p + 1 \leq 100 \): \[ 2p^2 - 2p - 99 \leq 0. \] Solving the quadratic \( 2p^2 - 2p - 99 = 0 \) using the quadratic formula: \[ p = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-99)}}{2 \cdot 2} = \frac{2 \pm \sqrt{4 + 792}}{4} = \frac{2 \pm \sqrt{796}}{4} = \frac{2 \pm 2\sqrt{199}}{4} = \frac{1 \pm \sqrt{199}}{2}. \] 7. **Finding the Roots**: Approximating \( \sqrt{199} \approx 14.1 \): - \( p_1 \approx \frac{1 - 14.1}{2} \approx -6.55 \) - \( p_2 \approx \frac{1 + 14.1}{2} \approx 7.55 \) Thus, the range for \( p \) is: \[ -6.55 < p < 7.55. \] 8. **Finding Integral Values of \( p \)**: The integral values of \( p \) in this range are: \[ -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7. \] This gives us a total of 14 integral values. 9. **Finding Corresponding Values of \( q \)**: Since \( q = p^2 \), for each integral \( p \), \( q \) will also be an integer. ### Conclusion: The number of integral values of \( (p, q) \) that satisfy the conditions is **14**.
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