NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119695
The radius of the circle \(2 x^2+2 y^2-3 x+2 y-1=0 \ldots . .\). units.
1 \(\frac{\sqrt{21}}{2}\)
2 \(\frac{\sqrt{21}}{4}\)
3 \(\frac{21}{4}\)
4 \(\frac{\sqrt{5}}{4}\)
Explanation:
B Given circle equation \(2 x^2+2 y^2-3 x+2 y-1=0\) \(x^2+y^2-\frac{3}{2} x+y-\frac{1}{2}=0\) By comparing above equation with general equation of circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(\mathrm{~g}=\frac{-3}{4}, \mathrm{f}=\frac{1}{2}, \quad \mathrm{c}=-\frac{1}{2}\) Radius \((\mathrm{R})=\sqrt{(\mathrm{g})^2+(\mathrm{f})^2-\mathrm{c}}\) \(=\sqrt{\left(\frac{-3}{4}\right)^2+\left(\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)}\) \(=\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}=\sqrt{\frac{21}{16}}\) \(R=\frac{\sqrt{21}}{4}\)
AP EAMCET-23.09.2020
Conic Section
119696
A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is
1 \(\mathrm{x}^2+\mathrm{y}^2=9\)
2 \(x^2+y^2=18\)
3 \(\mathrm{x}^2+\mathrm{y}^2=36\)
4 \(\mathrm{x}^2+\mathrm{y}^2=81\)
Explanation:
C Let the equation of a circle passing through one vertex of an equilateral triangle whose median of length \(3 a\) and having centre at origin is- \(x^2+y^2=4 a^2\) Given that, the length of the median is 9 units Hence, \(3 \mathrm{a}=9\) \(a=3\) Therefore the equation of the circle is- \(x^2+y^2=4 a^2\) \(x^2+y^2=4 \times(3)^2\) \(x^2+y^2=36\)
AP EAMCET-2017
Conic Section
119697
The point to which the origin is to be shifted to remove the first degree terms from the equation \(2 x^2+4 x y-6 y^2+2 x+8 y+1=0\) is
1 \(\left(\frac{7}{8}, \frac{3}{8}\right)\)
2 \(\left(\frac{-7}{8}, \frac{-3}{8}\right)\)
3 \(\left(\frac{-7}{8}, \frac{3}{8}\right)\)
4 \(\left(\frac{7}{8}, \frac{-3}{8}\right)\)
Explanation:
C Given equation - \(F=2 x^2+4 x y-6 y^2+2 x+8 y+1=0\) On partial differential we get \(\frac{\partial F}{\partial x}=4 x+4 y+2=0\) \(\frac{\partial F}{\partial y}=-12 y+4 x+8=0\) From solving equation (i) \& (ii), we get \(4 x+4 y+2=0\) \(4 x-12 y+8=0\) \(+\quad-\) \(16 y-6=0\) \(y=\frac{+6}{16}=\frac{3}{8}\) By putting the value of \(y\) in equation (i) we get \(4 \mathrm{x}+4 \times \frac{3}{8}+2=0\) \(4 \mathrm{x}=-\frac{7}{2} \Rightarrow \mathrm{x}=-\frac{7}{8}\)Hence \(\left(-\frac{7}{8}, \frac{3}{8}\right)\)
AP EAMCET-2017
Conic Section
119698
If a circle with radius 2.5 units passes through the points \((2,3)\) and \((5,7)\), then its centre is
1 \((1.5,2)\)
2 \((7,10)\)
3 \((3,4)\)
4 \((3.5,5)\)
Explanation:
D \(\mathrm{CA}=\mathrm{CB}\) \(\mathrm{CA}^2=\mathrm{CB}^2\) \(d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\) \(C A=\sqrt{(x-2)^2+(y-3)^2}=\sqrt{(x-5)^2+(y-2)^2}=C B\) \(C^2=C^2\) \((x-2)^2+(y-3)^2=(x-5)^2+(y-7)^2\) \(x^2+4-4 x+y^2+9-6 y=x^2+25-10 x+y^2+49-14 y\) \(6 x+8 y=61\)The point \((3.5,5)\) only satisfy the above equation (i) thus \((\mathrm{x}, \mathrm{y})=(3.5,5)\)
119695
The radius of the circle \(2 x^2+2 y^2-3 x+2 y-1=0 \ldots . .\). units.
1 \(\frac{\sqrt{21}}{2}\)
2 \(\frac{\sqrt{21}}{4}\)
3 \(\frac{21}{4}\)
4 \(\frac{\sqrt{5}}{4}\)
Explanation:
B Given circle equation \(2 x^2+2 y^2-3 x+2 y-1=0\) \(x^2+y^2-\frac{3}{2} x+y-\frac{1}{2}=0\) By comparing above equation with general equation of circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(\mathrm{~g}=\frac{-3}{4}, \mathrm{f}=\frac{1}{2}, \quad \mathrm{c}=-\frac{1}{2}\) Radius \((\mathrm{R})=\sqrt{(\mathrm{g})^2+(\mathrm{f})^2-\mathrm{c}}\) \(=\sqrt{\left(\frac{-3}{4}\right)^2+\left(\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)}\) \(=\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}=\sqrt{\frac{21}{16}}\) \(R=\frac{\sqrt{21}}{4}\)
AP EAMCET-23.09.2020
Conic Section
119696
A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is
1 \(\mathrm{x}^2+\mathrm{y}^2=9\)
2 \(x^2+y^2=18\)
3 \(\mathrm{x}^2+\mathrm{y}^2=36\)
4 \(\mathrm{x}^2+\mathrm{y}^2=81\)
Explanation:
C Let the equation of a circle passing through one vertex of an equilateral triangle whose median of length \(3 a\) and having centre at origin is- \(x^2+y^2=4 a^2\) Given that, the length of the median is 9 units Hence, \(3 \mathrm{a}=9\) \(a=3\) Therefore the equation of the circle is- \(x^2+y^2=4 a^2\) \(x^2+y^2=4 \times(3)^2\) \(x^2+y^2=36\)
AP EAMCET-2017
Conic Section
119697
The point to which the origin is to be shifted to remove the first degree terms from the equation \(2 x^2+4 x y-6 y^2+2 x+8 y+1=0\) is
1 \(\left(\frac{7}{8}, \frac{3}{8}\right)\)
2 \(\left(\frac{-7}{8}, \frac{-3}{8}\right)\)
3 \(\left(\frac{-7}{8}, \frac{3}{8}\right)\)
4 \(\left(\frac{7}{8}, \frac{-3}{8}\right)\)
Explanation:
C Given equation - \(F=2 x^2+4 x y-6 y^2+2 x+8 y+1=0\) On partial differential we get \(\frac{\partial F}{\partial x}=4 x+4 y+2=0\) \(\frac{\partial F}{\partial y}=-12 y+4 x+8=0\) From solving equation (i) \& (ii), we get \(4 x+4 y+2=0\) \(4 x-12 y+8=0\) \(+\quad-\) \(16 y-6=0\) \(y=\frac{+6}{16}=\frac{3}{8}\) By putting the value of \(y\) in equation (i) we get \(4 \mathrm{x}+4 \times \frac{3}{8}+2=0\) \(4 \mathrm{x}=-\frac{7}{2} \Rightarrow \mathrm{x}=-\frac{7}{8}\)Hence \(\left(-\frac{7}{8}, \frac{3}{8}\right)\)
AP EAMCET-2017
Conic Section
119698
If a circle with radius 2.5 units passes through the points \((2,3)\) and \((5,7)\), then its centre is
1 \((1.5,2)\)
2 \((7,10)\)
3 \((3,4)\)
4 \((3.5,5)\)
Explanation:
D \(\mathrm{CA}=\mathrm{CB}\) \(\mathrm{CA}^2=\mathrm{CB}^2\) \(d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\) \(C A=\sqrt{(x-2)^2+(y-3)^2}=\sqrt{(x-5)^2+(y-2)^2}=C B\) \(C^2=C^2\) \((x-2)^2+(y-3)^2=(x-5)^2+(y-7)^2\) \(x^2+4-4 x+y^2+9-6 y=x^2+25-10 x+y^2+49-14 y\) \(6 x+8 y=61\)The point \((3.5,5)\) only satisfy the above equation (i) thus \((\mathrm{x}, \mathrm{y})=(3.5,5)\)
119695
The radius of the circle \(2 x^2+2 y^2-3 x+2 y-1=0 \ldots . .\). units.
1 \(\frac{\sqrt{21}}{2}\)
2 \(\frac{\sqrt{21}}{4}\)
3 \(\frac{21}{4}\)
4 \(\frac{\sqrt{5}}{4}\)
Explanation:
B Given circle equation \(2 x^2+2 y^2-3 x+2 y-1=0\) \(x^2+y^2-\frac{3}{2} x+y-\frac{1}{2}=0\) By comparing above equation with general equation of circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(\mathrm{~g}=\frac{-3}{4}, \mathrm{f}=\frac{1}{2}, \quad \mathrm{c}=-\frac{1}{2}\) Radius \((\mathrm{R})=\sqrt{(\mathrm{g})^2+(\mathrm{f})^2-\mathrm{c}}\) \(=\sqrt{\left(\frac{-3}{4}\right)^2+\left(\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)}\) \(=\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}=\sqrt{\frac{21}{16}}\) \(R=\frac{\sqrt{21}}{4}\)
AP EAMCET-23.09.2020
Conic Section
119696
A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is
1 \(\mathrm{x}^2+\mathrm{y}^2=9\)
2 \(x^2+y^2=18\)
3 \(\mathrm{x}^2+\mathrm{y}^2=36\)
4 \(\mathrm{x}^2+\mathrm{y}^2=81\)
Explanation:
C Let the equation of a circle passing through one vertex of an equilateral triangle whose median of length \(3 a\) and having centre at origin is- \(x^2+y^2=4 a^2\) Given that, the length of the median is 9 units Hence, \(3 \mathrm{a}=9\) \(a=3\) Therefore the equation of the circle is- \(x^2+y^2=4 a^2\) \(x^2+y^2=4 \times(3)^2\) \(x^2+y^2=36\)
AP EAMCET-2017
Conic Section
119697
The point to which the origin is to be shifted to remove the first degree terms from the equation \(2 x^2+4 x y-6 y^2+2 x+8 y+1=0\) is
1 \(\left(\frac{7}{8}, \frac{3}{8}\right)\)
2 \(\left(\frac{-7}{8}, \frac{-3}{8}\right)\)
3 \(\left(\frac{-7}{8}, \frac{3}{8}\right)\)
4 \(\left(\frac{7}{8}, \frac{-3}{8}\right)\)
Explanation:
C Given equation - \(F=2 x^2+4 x y-6 y^2+2 x+8 y+1=0\) On partial differential we get \(\frac{\partial F}{\partial x}=4 x+4 y+2=0\) \(\frac{\partial F}{\partial y}=-12 y+4 x+8=0\) From solving equation (i) \& (ii), we get \(4 x+4 y+2=0\) \(4 x-12 y+8=0\) \(+\quad-\) \(16 y-6=0\) \(y=\frac{+6}{16}=\frac{3}{8}\) By putting the value of \(y\) in equation (i) we get \(4 \mathrm{x}+4 \times \frac{3}{8}+2=0\) \(4 \mathrm{x}=-\frac{7}{2} \Rightarrow \mathrm{x}=-\frac{7}{8}\)Hence \(\left(-\frac{7}{8}, \frac{3}{8}\right)\)
AP EAMCET-2017
Conic Section
119698
If a circle with radius 2.5 units passes through the points \((2,3)\) and \((5,7)\), then its centre is
1 \((1.5,2)\)
2 \((7,10)\)
3 \((3,4)\)
4 \((3.5,5)\)
Explanation:
D \(\mathrm{CA}=\mathrm{CB}\) \(\mathrm{CA}^2=\mathrm{CB}^2\) \(d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\) \(C A=\sqrt{(x-2)^2+(y-3)^2}=\sqrt{(x-5)^2+(y-2)^2}=C B\) \(C^2=C^2\) \((x-2)^2+(y-3)^2=(x-5)^2+(y-7)^2\) \(x^2+4-4 x+y^2+9-6 y=x^2+25-10 x+y^2+49-14 y\) \(6 x+8 y=61\)The point \((3.5,5)\) only satisfy the above equation (i) thus \((\mathrm{x}, \mathrm{y})=(3.5,5)\)
119695
The radius of the circle \(2 x^2+2 y^2-3 x+2 y-1=0 \ldots . .\). units.
1 \(\frac{\sqrt{21}}{2}\)
2 \(\frac{\sqrt{21}}{4}\)
3 \(\frac{21}{4}\)
4 \(\frac{\sqrt{5}}{4}\)
Explanation:
B Given circle equation \(2 x^2+2 y^2-3 x+2 y-1=0\) \(x^2+y^2-\frac{3}{2} x+y-\frac{1}{2}=0\) By comparing above equation with general equation of circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(\mathrm{~g}=\frac{-3}{4}, \mathrm{f}=\frac{1}{2}, \quad \mathrm{c}=-\frac{1}{2}\) Radius \((\mathrm{R})=\sqrt{(\mathrm{g})^2+(\mathrm{f})^2-\mathrm{c}}\) \(=\sqrt{\left(\frac{-3}{4}\right)^2+\left(\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)}\) \(=\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}=\sqrt{\frac{21}{16}}\) \(R=\frac{\sqrt{21}}{4}\)
AP EAMCET-23.09.2020
Conic Section
119696
A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is
1 \(\mathrm{x}^2+\mathrm{y}^2=9\)
2 \(x^2+y^2=18\)
3 \(\mathrm{x}^2+\mathrm{y}^2=36\)
4 \(\mathrm{x}^2+\mathrm{y}^2=81\)
Explanation:
C Let the equation of a circle passing through one vertex of an equilateral triangle whose median of length \(3 a\) and having centre at origin is- \(x^2+y^2=4 a^2\) Given that, the length of the median is 9 units Hence, \(3 \mathrm{a}=9\) \(a=3\) Therefore the equation of the circle is- \(x^2+y^2=4 a^2\) \(x^2+y^2=4 \times(3)^2\) \(x^2+y^2=36\)
AP EAMCET-2017
Conic Section
119697
The point to which the origin is to be shifted to remove the first degree terms from the equation \(2 x^2+4 x y-6 y^2+2 x+8 y+1=0\) is
1 \(\left(\frac{7}{8}, \frac{3}{8}\right)\)
2 \(\left(\frac{-7}{8}, \frac{-3}{8}\right)\)
3 \(\left(\frac{-7}{8}, \frac{3}{8}\right)\)
4 \(\left(\frac{7}{8}, \frac{-3}{8}\right)\)
Explanation:
C Given equation - \(F=2 x^2+4 x y-6 y^2+2 x+8 y+1=0\) On partial differential we get \(\frac{\partial F}{\partial x}=4 x+4 y+2=0\) \(\frac{\partial F}{\partial y}=-12 y+4 x+8=0\) From solving equation (i) \& (ii), we get \(4 x+4 y+2=0\) \(4 x-12 y+8=0\) \(+\quad-\) \(16 y-6=0\) \(y=\frac{+6}{16}=\frac{3}{8}\) By putting the value of \(y\) in equation (i) we get \(4 \mathrm{x}+4 \times \frac{3}{8}+2=0\) \(4 \mathrm{x}=-\frac{7}{2} \Rightarrow \mathrm{x}=-\frac{7}{8}\)Hence \(\left(-\frac{7}{8}, \frac{3}{8}\right)\)
AP EAMCET-2017
Conic Section
119698
If a circle with radius 2.5 units passes through the points \((2,3)\) and \((5,7)\), then its centre is
1 \((1.5,2)\)
2 \((7,10)\)
3 \((3,4)\)
4 \((3.5,5)\)
Explanation:
D \(\mathrm{CA}=\mathrm{CB}\) \(\mathrm{CA}^2=\mathrm{CB}^2\) \(d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\) \(C A=\sqrt{(x-2)^2+(y-3)^2}=\sqrt{(x-5)^2+(y-2)^2}=C B\) \(C^2=C^2\) \((x-2)^2+(y-3)^2=(x-5)^2+(y-7)^2\) \(x^2+4-4 x+y^2+9-6 y=x^2+25-10 x+y^2+49-14 y\) \(6 x+8 y=61\)The point \((3.5,5)\) only satisfy the above equation (i) thus \((\mathrm{x}, \mathrm{y})=(3.5,5)\)
