A) -2 |
B) 2 |
C) 1 |
D) 0 |
B) 2 |
Let the general equation of circle be
\( \Large x^{2}+y^{2}+2gx+2fy+c=0 \)the equation of circle passing through \( \Large \left(0,\ 0\right),\ \left(2,\ 0\right)\ and\ \left(0,\ -2\right) \)
\( \Large c = 0 \) ,,,(i)
\( \Large 4 + 4g + c = 0 \) ...(ii)
and \( \Large 4- 4f + c = 0 \) ,,,(iii)
Solving equations (i), (ii) and (iii), we get
c=0,g=-1,f=1 ...(iv)
The equation of circle becomes \( \Large x^{2}+y^{2}-2x+2y=0 \)
since, it is passes through \( \Large \left(k,\ -2\right) \), we get
\( \Large k^{2}+4-2k-4=0 => k = 0,\ 2 \)
We have already take a point \( \Large \left( 0, -2 \right) \) so we take only k = 2.
1). The co-ordinate axis rotated through an angle \( \Large 135 ^{\circ} \). If the co-ordinates of a points P in the new system are known to be \( \Large \left(4,\ -3\right) \), then the co-ordinates of P in the original systems are:
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2). Area of quadrilateral whose vertices are \( \Large \left(2,\ 3\right),\ \left(3,\ 4\right),\ \left(4,\ 5\right)\ and\ \left(5,\ 6\right) \)
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3). Co-ordinates of the foot of the perpendicular drawn from \( \Large \left(0,\ 0\right) \) to the line joining \( \Large \left(a \cos \alpha ,\ a \sin \alpha\right)\ and\ \left(a \cos \beta ,\ a \sin \beta \right) \) are:
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4). An equilateral triangle has each side equal to a the co-ordinates of its vertices are \( \Large \left(x_{1},\ y_{1}\right),\ \left(x_{2},\ y_{2}\right)\ and\ \left(x_{3},\ y_{3}\right) \), then the square of determinant \( \begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix} \) equals
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5). The incentre of triangle with vertices \( \Large \left(1,\ \sqrt{3}\right),\ \left(0,\ 0\right)\ and\ \left(2,\ 0\right) \) is:
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6). If \( \Large P \left(a_{1},\ b_{1}\right)\ and\ Q \left(a_{2},\ b_{2}\right) \) are two points, the \( \Large OP.OQ \cos \left( \angle POQ\right) \) is (O is origin ):
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7). If a point \( \Large P \left(4,\ 3\right) \) is shifted by a distance \( \Large \sqrt{2} \) unit parallel to the line y = x, then co-ordinates of P in new position are:
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8). The orthocentre of a triangle formed by lines \( \Large 2x+y=2,\ x-2y=1\ and\ x+y=1 \) is:
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9). The incentre of triangle formed by lines x = 0, y = 0 and 3x+4y = 12 is at:
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10). If the vertices of a triangle have integral co ordinates, the triangle cannot be:
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