NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119994
The length of the common chord of the circles \(x^2+y^2+3 x+5 y+4=0\) and \(x^2+y^2+5 x+3 y+4\) \(=0\) is \(\qquad\) units.
1 3
2 2
3 6
4 4
Explanation:
D Given, \(\mathrm{C}_1: \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+5 \mathrm{y}+4=0\) \(C_2: x^2+y^2+5 x+3 y+4=0\) On solving these two circles, we get - \(\mathrm{x}=\mathrm{y}\) Substituting \(x=y\) in equation (i), we get - \(2 \mathrm{x}^2+8 \mathrm{x}+4=0\) \(\mathrm{x}^2+4 \mathrm{x}+2=0\) \((\mathrm{x}+2)^2=2\) \(\mathrm{x}+2= \pm \sqrt{2}, \quad \mathrm{x}_{1,2}=-2 \pm \sqrt{2}\) \(\mathrm{x}=\mathrm{y}=\mathrm{y}_{1,2}=-2 \pm \sqrt{2}\) Length \(=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}=\sqrt{2}\left(\mathrm{x}_1-\mathrm{x}_2\right)\) \(=\sqrt{2}[(-2+\sqrt{2})-(-2-\sqrt{2})]=4\)
AP EAPCET-25.08.2021
Conic Section
119995
Find the equation of the circle which passes through the point \((1,2)\) and the points of intersection of the circles \(x^2+y^2-8 x-6 y+21=\) 0 and \(x^2+y^2-2 x-15=0\)
119996
The perpendicular distance from the point \((1,2)\) to common chord of the circles \(x^2+y^2-2 x\) \(+4 y-4=0\) and \(x^2+y^2+4 x-6 y-3=0\) is \(\qquad\) units.
119997
The radical centre of the circles \(x^2+y^2+3 x+\) \(2 y+1=0, x^2+y^2-x+6 y+5=0\) and \(x^2+y^2+\) \(5 x-8 y+15=0\) is
1 \((3,2)\)
2 \((-3,-2)\)
3 \((2,3)\)
4 \((-2,-3)\)
Explanation:
A Given, \(\mathrm{S}_1: \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+2 \mathrm{y}+1=0\) \(\mathrm{S}_2: \mathrm{x}^2+\mathrm{y}^2-\mathrm{x}+6 \mathrm{y}+5=0\) \(\mathrm{S}_3: \mathrm{x}^2+\mathrm{y}^2+5 \mathrm{x}-8 \mathrm{y}+15=0\) Let, \((\mathrm{x}, \mathrm{y})\) be the radical centre, then \(\mathrm{S}_1-\mathrm{S}_2=0\) \(\mathrm{S}_2-\mathrm{S}_3=0\) \(4 \mathrm{x}-4 \mathrm{y}-4=0\) and \(-6 \mathrm{x}+14 \mathrm{y}-10=0\) \(\mathrm{x}=\mathrm{y}+1\) and \(3 \mathrm{x}-7 \mathrm{y}+5=0\) \(3(\mathrm{y}+1)-7 \mathrm{y}+5=0\) \(3 \mathrm{y}+3-7 \mathrm{y}+5=0\) \(-4 \mathrm{y}+8=0\) \(\therefore \quad \mathrm{y}=2\) and \(\mathrm{x}=2+1=3\) Radical centre is \((3,2)\).
119994
The length of the common chord of the circles \(x^2+y^2+3 x+5 y+4=0\) and \(x^2+y^2+5 x+3 y+4\) \(=0\) is \(\qquad\) units.
1 3
2 2
3 6
4 4
Explanation:
D Given, \(\mathrm{C}_1: \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+5 \mathrm{y}+4=0\) \(C_2: x^2+y^2+5 x+3 y+4=0\) On solving these two circles, we get - \(\mathrm{x}=\mathrm{y}\) Substituting \(x=y\) in equation (i), we get - \(2 \mathrm{x}^2+8 \mathrm{x}+4=0\) \(\mathrm{x}^2+4 \mathrm{x}+2=0\) \((\mathrm{x}+2)^2=2\) \(\mathrm{x}+2= \pm \sqrt{2}, \quad \mathrm{x}_{1,2}=-2 \pm \sqrt{2}\) \(\mathrm{x}=\mathrm{y}=\mathrm{y}_{1,2}=-2 \pm \sqrt{2}\) Length \(=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}=\sqrt{2}\left(\mathrm{x}_1-\mathrm{x}_2\right)\) \(=\sqrt{2}[(-2+\sqrt{2})-(-2-\sqrt{2})]=4\)
AP EAPCET-25.08.2021
Conic Section
119995
Find the equation of the circle which passes through the point \((1,2)\) and the points of intersection of the circles \(x^2+y^2-8 x-6 y+21=\) 0 and \(x^2+y^2-2 x-15=0\)
119996
The perpendicular distance from the point \((1,2)\) to common chord of the circles \(x^2+y^2-2 x\) \(+4 y-4=0\) and \(x^2+y^2+4 x-6 y-3=0\) is \(\qquad\) units.
119997
The radical centre of the circles \(x^2+y^2+3 x+\) \(2 y+1=0, x^2+y^2-x+6 y+5=0\) and \(x^2+y^2+\) \(5 x-8 y+15=0\) is
1 \((3,2)\)
2 \((-3,-2)\)
3 \((2,3)\)
4 \((-2,-3)\)
Explanation:
A Given, \(\mathrm{S}_1: \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+2 \mathrm{y}+1=0\) \(\mathrm{S}_2: \mathrm{x}^2+\mathrm{y}^2-\mathrm{x}+6 \mathrm{y}+5=0\) \(\mathrm{S}_3: \mathrm{x}^2+\mathrm{y}^2+5 \mathrm{x}-8 \mathrm{y}+15=0\) Let, \((\mathrm{x}, \mathrm{y})\) be the radical centre, then \(\mathrm{S}_1-\mathrm{S}_2=0\) \(\mathrm{S}_2-\mathrm{S}_3=0\) \(4 \mathrm{x}-4 \mathrm{y}-4=0\) and \(-6 \mathrm{x}+14 \mathrm{y}-10=0\) \(\mathrm{x}=\mathrm{y}+1\) and \(3 \mathrm{x}-7 \mathrm{y}+5=0\) \(3(\mathrm{y}+1)-7 \mathrm{y}+5=0\) \(3 \mathrm{y}+3-7 \mathrm{y}+5=0\) \(-4 \mathrm{y}+8=0\) \(\therefore \quad \mathrm{y}=2\) and \(\mathrm{x}=2+1=3\) Radical centre is \((3,2)\).
119994
The length of the common chord of the circles \(x^2+y^2+3 x+5 y+4=0\) and \(x^2+y^2+5 x+3 y+4\) \(=0\) is \(\qquad\) units.
1 3
2 2
3 6
4 4
Explanation:
D Given, \(\mathrm{C}_1: \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+5 \mathrm{y}+4=0\) \(C_2: x^2+y^2+5 x+3 y+4=0\) On solving these two circles, we get - \(\mathrm{x}=\mathrm{y}\) Substituting \(x=y\) in equation (i), we get - \(2 \mathrm{x}^2+8 \mathrm{x}+4=0\) \(\mathrm{x}^2+4 \mathrm{x}+2=0\) \((\mathrm{x}+2)^2=2\) \(\mathrm{x}+2= \pm \sqrt{2}, \quad \mathrm{x}_{1,2}=-2 \pm \sqrt{2}\) \(\mathrm{x}=\mathrm{y}=\mathrm{y}_{1,2}=-2 \pm \sqrt{2}\) Length \(=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}=\sqrt{2}\left(\mathrm{x}_1-\mathrm{x}_2\right)\) \(=\sqrt{2}[(-2+\sqrt{2})-(-2-\sqrt{2})]=4\)
AP EAPCET-25.08.2021
Conic Section
119995
Find the equation of the circle which passes through the point \((1,2)\) and the points of intersection of the circles \(x^2+y^2-8 x-6 y+21=\) 0 and \(x^2+y^2-2 x-15=0\)
119996
The perpendicular distance from the point \((1,2)\) to common chord of the circles \(x^2+y^2-2 x\) \(+4 y-4=0\) and \(x^2+y^2+4 x-6 y-3=0\) is \(\qquad\) units.
119997
The radical centre of the circles \(x^2+y^2+3 x+\) \(2 y+1=0, x^2+y^2-x+6 y+5=0\) and \(x^2+y^2+\) \(5 x-8 y+15=0\) is
1 \((3,2)\)
2 \((-3,-2)\)
3 \((2,3)\)
4 \((-2,-3)\)
Explanation:
A Given, \(\mathrm{S}_1: \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+2 \mathrm{y}+1=0\) \(\mathrm{S}_2: \mathrm{x}^2+\mathrm{y}^2-\mathrm{x}+6 \mathrm{y}+5=0\) \(\mathrm{S}_3: \mathrm{x}^2+\mathrm{y}^2+5 \mathrm{x}-8 \mathrm{y}+15=0\) Let, \((\mathrm{x}, \mathrm{y})\) be the radical centre, then \(\mathrm{S}_1-\mathrm{S}_2=0\) \(\mathrm{S}_2-\mathrm{S}_3=0\) \(4 \mathrm{x}-4 \mathrm{y}-4=0\) and \(-6 \mathrm{x}+14 \mathrm{y}-10=0\) \(\mathrm{x}=\mathrm{y}+1\) and \(3 \mathrm{x}-7 \mathrm{y}+5=0\) \(3(\mathrm{y}+1)-7 \mathrm{y}+5=0\) \(3 \mathrm{y}+3-7 \mathrm{y}+5=0\) \(-4 \mathrm{y}+8=0\) \(\therefore \quad \mathrm{y}=2\) and \(\mathrm{x}=2+1=3\) Radical centre is \((3,2)\).
119994
The length of the common chord of the circles \(x^2+y^2+3 x+5 y+4=0\) and \(x^2+y^2+5 x+3 y+4\) \(=0\) is \(\qquad\) units.
1 3
2 2
3 6
4 4
Explanation:
D Given, \(\mathrm{C}_1: \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+5 \mathrm{y}+4=0\) \(C_2: x^2+y^2+5 x+3 y+4=0\) On solving these two circles, we get - \(\mathrm{x}=\mathrm{y}\) Substituting \(x=y\) in equation (i), we get - \(2 \mathrm{x}^2+8 \mathrm{x}+4=0\) \(\mathrm{x}^2+4 \mathrm{x}+2=0\) \((\mathrm{x}+2)^2=2\) \(\mathrm{x}+2= \pm \sqrt{2}, \quad \mathrm{x}_{1,2}=-2 \pm \sqrt{2}\) \(\mathrm{x}=\mathrm{y}=\mathrm{y}_{1,2}=-2 \pm \sqrt{2}\) Length \(=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}=\sqrt{2}\left(\mathrm{x}_1-\mathrm{x}_2\right)\) \(=\sqrt{2}[(-2+\sqrt{2})-(-2-\sqrt{2})]=4\)
AP EAPCET-25.08.2021
Conic Section
119995
Find the equation of the circle which passes through the point \((1,2)\) and the points of intersection of the circles \(x^2+y^2-8 x-6 y+21=\) 0 and \(x^2+y^2-2 x-15=0\)
119996
The perpendicular distance from the point \((1,2)\) to common chord of the circles \(x^2+y^2-2 x\) \(+4 y-4=0\) and \(x^2+y^2+4 x-6 y-3=0\) is \(\qquad\) units.
119997
The radical centre of the circles \(x^2+y^2+3 x+\) \(2 y+1=0, x^2+y^2-x+6 y+5=0\) and \(x^2+y^2+\) \(5 x-8 y+15=0\) is
1 \((3,2)\)
2 \((-3,-2)\)
3 \((2,3)\)
4 \((-2,-3)\)
Explanation:
A Given, \(\mathrm{S}_1: \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+2 \mathrm{y}+1=0\) \(\mathrm{S}_2: \mathrm{x}^2+\mathrm{y}^2-\mathrm{x}+6 \mathrm{y}+5=0\) \(\mathrm{S}_3: \mathrm{x}^2+\mathrm{y}^2+5 \mathrm{x}-8 \mathrm{y}+15=0\) Let, \((\mathrm{x}, \mathrm{y})\) be the radical centre, then \(\mathrm{S}_1-\mathrm{S}_2=0\) \(\mathrm{S}_2-\mathrm{S}_3=0\) \(4 \mathrm{x}-4 \mathrm{y}-4=0\) and \(-6 \mathrm{x}+14 \mathrm{y}-10=0\) \(\mathrm{x}=\mathrm{y}+1\) and \(3 \mathrm{x}-7 \mathrm{y}+5=0\) \(3(\mathrm{y}+1)-7 \mathrm{y}+5=0\) \(3 \mathrm{y}+3-7 \mathrm{y}+5=0\) \(-4 \mathrm{y}+8=0\) \(\therefore \quad \mathrm{y}=2\) and \(\mathrm{x}=2+1=3\) Radical centre is \((3,2)\).
