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A tangent is drawn to the circle 2(x^(...

A tangent is drawn to the circle `2(x^(2)+y^(2))-3x+4y=0` and it touches the circle at point A. If the tangent passes through the point P(2, 1),then PA=

A

4

B

2

C

`2sqrt(2)`

D

none of these

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To solve the problem, we need to find the length of the segment PA, where P is the point (2, 1) and A is the point where the tangent touches the circle given by the equation \( 2(x^2 + y^2) - 3x + 4y = 0 \). ### Step 1: Rewrite the Circle Equation First, we simplify the equation of the circle: \[ 2(x^2 + y^2) - 3x + 4y = 0 \] Dividing the entire equation by 2 gives: \[ x^2 + y^2 - \frac{3}{2}x + 2y = 0 \] ### Step 2: Identify the Center and Radius The general form of a circle is given by: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] From our equation, we can identify: - \( g = -\frac{3}{4} \) - \( f = 1 \) - \( c = 0 \) The center of the circle \((h, k)\) can be calculated as: \[ h = -g = \frac{3}{4}, \quad k = -f = -1 \] Thus, the center of the circle is: \[ \left(\frac{3}{4}, -1\right) \] ### Step 3: Calculate the Radius The radius \( r \) of the circle can be calculated using the formula: \[ r = \sqrt{h^2 + k^2 - c} \] Substituting the values: \[ r = \sqrt{\left(\frac{3}{4}\right)^2 + (-1)^2 - 0} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{9}{16} + \frac{16}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \] ### Step 4: Calculate the Distance from Point P to the Center Now we calculate the distance from point P(2, 1) to the center of the circle \(\left(\frac{3}{4}, -1\right)\): \[ \text{Distance} = \sqrt{\left(2 - \frac{3}{4}\right)^2 + \left(1 - (-1)\right)^2} \] Calculating the x-component: \[ 2 - \frac{3}{4} = \frac{8}{4} - \frac{3}{4} = \frac{5}{4} \] Calculating the y-component: \[ 1 - (-1) = 1 + 1 = 2 \] Now substituting back into the distance formula: \[ \text{Distance} = \sqrt{\left(\frac{5}{4}\right)^2 + 2^2} = \sqrt{\frac{25}{16} + 4} = \sqrt{\frac{25}{16} + \frac{64}{16}} = \sqrt{\frac{89}{16}} = \frac{\sqrt{89}}{4} \] ### Step 5: Use the Tangent Length Formula The length of the tangent from point P to the circle can be calculated using the formula: \[ PA = \sqrt{d^2 - r^2} \] Where \( d \) is the distance from point P to the center and \( r \) is the radius: \[ PA = \sqrt{\left(\frac{\sqrt{89}}{4}\right)^2 - \left(\frac{5}{4}\right)^2} \] Calculating: \[ PA = \sqrt{\frac{89}{16} - \frac{25}{16}} = \sqrt{\frac{64}{16}} = \sqrt{4} = 2 \] ### Final Answer Thus, the length \( PA \) is: \[ \boxed{2} \]
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OBJECTIVE RD SHARMA ENGLISH-CIRCLES-Chapter Test
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  2. Two tangents to the circle x^(2) +y^(2) = 4 at the points A and B meet...

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  3. A tangent is drawn to the circle 2(x^(2)+y^(2))-3x+4y=0 and it touch...

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  4. the length of the chord of the circle x^(2)+y^(2)=25 passing through ...

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  5. If the points A(2, 5) and B are symmetrical about the tangent to the c...

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  6. The equation of the circle of radius 2 sqrt(2) whose centre lies on th...

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  7. Prove that the maximum number of points with rational coordinates on a...

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  8. The equation of a circle C is x^(2)+y^(2)-6x-8y-11=0. The number of re...

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  9. Two circles, each of radius 5, have a common tangent at (1, 1) whose e...

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  10. The number of points on the circle 2(x^(2)+y^(2))=3x which are at a di...

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  11. A ray of light incident at the point ( -2,-1) gets reflected from the ...

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  12. The point on the straight line y = 2x + 11 which is nearest to the cir...

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  13. Extremities of a diagonal of a rectangle are (0, 0) and (4, 3). The eq...

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  14. The equation of the circle which has a tangent 2x-y-1= 0 at P( 3,5) on...

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  15. The angle of intersection of the circles x^(2)+y^(2)=4 and x^(2)+y^(2...

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  16. The normal at the point (3,4) on a circle cuts the circle at the poins...

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  17. The inverse point of (1, -1) with respect to x^(2)+y^(2)=4, is

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  18. A variable circle passes through the point A(a,b) and touches the x-a...

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  19. The radius of the circle r^(2)-2sqrt(2r) (cos theta + sin theta)-5=0, ...

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  20. A straight rot of length 9 units slides with its ends A,B always on th...

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