NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
120061
The radical centre of the system of circles, \(x^2+y^2+4 x+7=0,\) \(2\left(x^2+y^2\right)+3 x+5 y+9=0\) and \(x^2+y^2+y=0\) is
1 \((-2,-1)\)
2 \((1,-2)\)
3 \((-1,-2)\)
4 None of these
Explanation:
A We have:- The system of circles, \(x^2+y^2+4 x+7=0\) \(2\left(x^2+y^2\right)+3 x+5 y+9=0\) \(x^2+y^2+\frac{3}{2} x+\frac{5}{2} y+\frac{9}{2}=0\) \(x^2+y^2+y=0\) On subtracting \(\mathrm{Eq}^{\mathrm{n}}\) (ii) from \(\mathrm{Eq}^{\mathrm{n}}\) (i), we get \(4 x-\frac{3}{2} x-\frac{5}{2} y+7-\frac{9}{2}=0\) \(\frac{5}{2} x-\frac{5}{2} y+\frac{5}{2}=0\) \(x-y+1=0\) On subtracting \(\mathrm{Eq}^{\mathrm{n}}\) (iii) from \(\mathrm{Eq}^{\mathrm{n}}\) (i), we get \(\Rightarrow \quad\left(\mathrm{x}^2+\mathrm{y}^2+\mathrm{y}\right)-\left(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}+7\right)=0\) \(4 \mathrm{x}-\mathrm{y}+7=0\) On solving Eq \({ }^n\) (iv) and (v), we get:- \(x=-2, \quad y=-1\)Hence, radical centre is \((-2,-1)\).
Manipal UGET-2020
Conic Section
120062
If the lengths of the tangents drawn from a point \(P\) to the three circles \(\mathrm{x}^2+\mathrm{y}^2-\mathbf{4}=\mathbf{0}\), \(x^2+y^2-2 x+3 y=0\) and \(x^2+y^2+7 y-18=0\) are equal, then the coordinates of \(P\) are
1 \((2,5)\)
2 \((3,4)\)
3 \((4,3)\)
4 \((5,2)\)
Explanation:
D Given equation of circle - \(x^2+y^2-4=0\) \(x^2+y^2-2 x+3 y=0\) \(\mathrm{x}^2+\mathrm{y}^2+7 \mathrm{y}-18=0\) Radical axis of the circle by equation (i) and (iii), we get- \(-7 y+14=0\) \(y=2\) By equation (i) and (ii) Radical axis of the circle \(2 x-3 y-4=0\) From equation (iv) and (v) \(2 \mathrm{x}-6-4=0\) \(\mathrm{x}=5\)Radical center \((5,2)=\mathrm{P}=(5,2)\)
AP EAMCET-08.07.2022
Conic Section
120063
The centre of the circle passing through the points of intersection of the circles \((x+3)^2+(y\) \(+2)^2=25\) and \((x-2)^2+(y-3)^2=25\) and cutting the circle \((x+1)^2+(y-2)^2=16\) orthogonally is
1 \(\left(\frac{-27}{2}, \frac{-25}{2}\right)\)
2 \((0,0)\)
3 \(\left(\frac{16}{3}, \frac{-25}{4}\right)\)
4 \(\left(\frac{4}{7}, \frac{3}{7}\right)\)
Explanation:
A Equation of circle passes through the point of intersections of given circles, \(\qquad \mathrm{S}_1 \equiv(\mathrm{x}+3)^2+(\mathrm{y}+2)^2=25\) \(\text { And, } \quad \mathrm{S}_2 \equiv(\mathrm{x}-2)^2+(\mathrm{y}-3)^2=25\) \(\text { is } \mathrm{S}_1+\lambda \mathrm{S}_2=0\) \((1+\lambda) \mathrm{x}^2+(1+\lambda) \mathrm{y}^2+(6-4 \lambda) \mathrm{x}+(4-6 \lambda) \mathrm{y}\) \(-12(1+\lambda)=0\) \(\mathrm{x}^2+\mathrm{y}^2+2\left(\frac{3-2 \lambda}{1+\lambda}\right) \mathrm{x}+2\left(\frac{2-3 \lambda}{1+\lambda}\right) \mathrm{y}-12=0\) And, Cuts the another given circle \((x+1)^2+(y-2)^2=16\) Orthogonally, so \(2\left(\frac{3-2 \lambda}{1+\lambda}\right)-4\left(\frac{2-3 \lambda}{1+\lambda}\right)=-12-11\) \(\Rightarrow 6-4 \lambda-8+12 \lambda=-23-23 \lambda\) \(\Rightarrow 31 \lambda=-21 \Rightarrow \lambda=-\frac{21}{31}\) \(\therefore\) Centre of the circle (i) is \(\left(-\frac{3+\frac{42}{31}}{1-\frac{21}{31}},-\frac{2+\frac{63}{31}}{1-\frac{21}{31}}\right)=\left(-\frac{135}{10},-\frac{125}{10}\right)=\left(-\frac{27}{2},-\frac{25}{2}\right)\)
TS EAMCET-11.09.2020
Conic Section
120064
If the radical axis of the circles \(x^2+y^2+2 \alpha x+2 \beta y+c=0\) and \(x^2+y^2+\frac{3}{2} x+4 y+c=0\) touches the circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}+2 \mathrm{y}+\mathbf{1}=\mathbf{0}\), then \(4 \alpha \beta-8 \alpha-3 \beta+10=\)
120061
The radical centre of the system of circles, \(x^2+y^2+4 x+7=0,\) \(2\left(x^2+y^2\right)+3 x+5 y+9=0\) and \(x^2+y^2+y=0\) is
1 \((-2,-1)\)
2 \((1,-2)\)
3 \((-1,-2)\)
4 None of these
Explanation:
A We have:- The system of circles, \(x^2+y^2+4 x+7=0\) \(2\left(x^2+y^2\right)+3 x+5 y+9=0\) \(x^2+y^2+\frac{3}{2} x+\frac{5}{2} y+\frac{9}{2}=0\) \(x^2+y^2+y=0\) On subtracting \(\mathrm{Eq}^{\mathrm{n}}\) (ii) from \(\mathrm{Eq}^{\mathrm{n}}\) (i), we get \(4 x-\frac{3}{2} x-\frac{5}{2} y+7-\frac{9}{2}=0\) \(\frac{5}{2} x-\frac{5}{2} y+\frac{5}{2}=0\) \(x-y+1=0\) On subtracting \(\mathrm{Eq}^{\mathrm{n}}\) (iii) from \(\mathrm{Eq}^{\mathrm{n}}\) (i), we get \(\Rightarrow \quad\left(\mathrm{x}^2+\mathrm{y}^2+\mathrm{y}\right)-\left(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}+7\right)=0\) \(4 \mathrm{x}-\mathrm{y}+7=0\) On solving Eq \({ }^n\) (iv) and (v), we get:- \(x=-2, \quad y=-1\)Hence, radical centre is \((-2,-1)\).
Manipal UGET-2020
Conic Section
120062
If the lengths of the tangents drawn from a point \(P\) to the three circles \(\mathrm{x}^2+\mathrm{y}^2-\mathbf{4}=\mathbf{0}\), \(x^2+y^2-2 x+3 y=0\) and \(x^2+y^2+7 y-18=0\) are equal, then the coordinates of \(P\) are
1 \((2,5)\)
2 \((3,4)\)
3 \((4,3)\)
4 \((5,2)\)
Explanation:
D Given equation of circle - \(x^2+y^2-4=0\) \(x^2+y^2-2 x+3 y=0\) \(\mathrm{x}^2+\mathrm{y}^2+7 \mathrm{y}-18=0\) Radical axis of the circle by equation (i) and (iii), we get- \(-7 y+14=0\) \(y=2\) By equation (i) and (ii) Radical axis of the circle \(2 x-3 y-4=0\) From equation (iv) and (v) \(2 \mathrm{x}-6-4=0\) \(\mathrm{x}=5\)Radical center \((5,2)=\mathrm{P}=(5,2)\)
AP EAMCET-08.07.2022
Conic Section
120063
The centre of the circle passing through the points of intersection of the circles \((x+3)^2+(y\) \(+2)^2=25\) and \((x-2)^2+(y-3)^2=25\) and cutting the circle \((x+1)^2+(y-2)^2=16\) orthogonally is
1 \(\left(\frac{-27}{2}, \frac{-25}{2}\right)\)
2 \((0,0)\)
3 \(\left(\frac{16}{3}, \frac{-25}{4}\right)\)
4 \(\left(\frac{4}{7}, \frac{3}{7}\right)\)
Explanation:
A Equation of circle passes through the point of intersections of given circles, \(\qquad \mathrm{S}_1 \equiv(\mathrm{x}+3)^2+(\mathrm{y}+2)^2=25\) \(\text { And, } \quad \mathrm{S}_2 \equiv(\mathrm{x}-2)^2+(\mathrm{y}-3)^2=25\) \(\text { is } \mathrm{S}_1+\lambda \mathrm{S}_2=0\) \((1+\lambda) \mathrm{x}^2+(1+\lambda) \mathrm{y}^2+(6-4 \lambda) \mathrm{x}+(4-6 \lambda) \mathrm{y}\) \(-12(1+\lambda)=0\) \(\mathrm{x}^2+\mathrm{y}^2+2\left(\frac{3-2 \lambda}{1+\lambda}\right) \mathrm{x}+2\left(\frac{2-3 \lambda}{1+\lambda}\right) \mathrm{y}-12=0\) And, Cuts the another given circle \((x+1)^2+(y-2)^2=16\) Orthogonally, so \(2\left(\frac{3-2 \lambda}{1+\lambda}\right)-4\left(\frac{2-3 \lambda}{1+\lambda}\right)=-12-11\) \(\Rightarrow 6-4 \lambda-8+12 \lambda=-23-23 \lambda\) \(\Rightarrow 31 \lambda=-21 \Rightarrow \lambda=-\frac{21}{31}\) \(\therefore\) Centre of the circle (i) is \(\left(-\frac{3+\frac{42}{31}}{1-\frac{21}{31}},-\frac{2+\frac{63}{31}}{1-\frac{21}{31}}\right)=\left(-\frac{135}{10},-\frac{125}{10}\right)=\left(-\frac{27}{2},-\frac{25}{2}\right)\)
TS EAMCET-11.09.2020
Conic Section
120064
If the radical axis of the circles \(x^2+y^2+2 \alpha x+2 \beta y+c=0\) and \(x^2+y^2+\frac{3}{2} x+4 y+c=0\) touches the circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}+2 \mathrm{y}+\mathbf{1}=\mathbf{0}\), then \(4 \alpha \beta-8 \alpha-3 \beta+10=\)
120061
The radical centre of the system of circles, \(x^2+y^2+4 x+7=0,\) \(2\left(x^2+y^2\right)+3 x+5 y+9=0\) and \(x^2+y^2+y=0\) is
1 \((-2,-1)\)
2 \((1,-2)\)
3 \((-1,-2)\)
4 None of these
Explanation:
A We have:- The system of circles, \(x^2+y^2+4 x+7=0\) \(2\left(x^2+y^2\right)+3 x+5 y+9=0\) \(x^2+y^2+\frac{3}{2} x+\frac{5}{2} y+\frac{9}{2}=0\) \(x^2+y^2+y=0\) On subtracting \(\mathrm{Eq}^{\mathrm{n}}\) (ii) from \(\mathrm{Eq}^{\mathrm{n}}\) (i), we get \(4 x-\frac{3}{2} x-\frac{5}{2} y+7-\frac{9}{2}=0\) \(\frac{5}{2} x-\frac{5}{2} y+\frac{5}{2}=0\) \(x-y+1=0\) On subtracting \(\mathrm{Eq}^{\mathrm{n}}\) (iii) from \(\mathrm{Eq}^{\mathrm{n}}\) (i), we get \(\Rightarrow \quad\left(\mathrm{x}^2+\mathrm{y}^2+\mathrm{y}\right)-\left(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}+7\right)=0\) \(4 \mathrm{x}-\mathrm{y}+7=0\) On solving Eq \({ }^n\) (iv) and (v), we get:- \(x=-2, \quad y=-1\)Hence, radical centre is \((-2,-1)\).
Manipal UGET-2020
Conic Section
120062
If the lengths of the tangents drawn from a point \(P\) to the three circles \(\mathrm{x}^2+\mathrm{y}^2-\mathbf{4}=\mathbf{0}\), \(x^2+y^2-2 x+3 y=0\) and \(x^2+y^2+7 y-18=0\) are equal, then the coordinates of \(P\) are
1 \((2,5)\)
2 \((3,4)\)
3 \((4,3)\)
4 \((5,2)\)
Explanation:
D Given equation of circle - \(x^2+y^2-4=0\) \(x^2+y^2-2 x+3 y=0\) \(\mathrm{x}^2+\mathrm{y}^2+7 \mathrm{y}-18=0\) Radical axis of the circle by equation (i) and (iii), we get- \(-7 y+14=0\) \(y=2\) By equation (i) and (ii) Radical axis of the circle \(2 x-3 y-4=0\) From equation (iv) and (v) \(2 \mathrm{x}-6-4=0\) \(\mathrm{x}=5\)Radical center \((5,2)=\mathrm{P}=(5,2)\)
AP EAMCET-08.07.2022
Conic Section
120063
The centre of the circle passing through the points of intersection of the circles \((x+3)^2+(y\) \(+2)^2=25\) and \((x-2)^2+(y-3)^2=25\) and cutting the circle \((x+1)^2+(y-2)^2=16\) orthogonally is
1 \(\left(\frac{-27}{2}, \frac{-25}{2}\right)\)
2 \((0,0)\)
3 \(\left(\frac{16}{3}, \frac{-25}{4}\right)\)
4 \(\left(\frac{4}{7}, \frac{3}{7}\right)\)
Explanation:
A Equation of circle passes through the point of intersections of given circles, \(\qquad \mathrm{S}_1 \equiv(\mathrm{x}+3)^2+(\mathrm{y}+2)^2=25\) \(\text { And, } \quad \mathrm{S}_2 \equiv(\mathrm{x}-2)^2+(\mathrm{y}-3)^2=25\) \(\text { is } \mathrm{S}_1+\lambda \mathrm{S}_2=0\) \((1+\lambda) \mathrm{x}^2+(1+\lambda) \mathrm{y}^2+(6-4 \lambda) \mathrm{x}+(4-6 \lambda) \mathrm{y}\) \(-12(1+\lambda)=0\) \(\mathrm{x}^2+\mathrm{y}^2+2\left(\frac{3-2 \lambda}{1+\lambda}\right) \mathrm{x}+2\left(\frac{2-3 \lambda}{1+\lambda}\right) \mathrm{y}-12=0\) And, Cuts the another given circle \((x+1)^2+(y-2)^2=16\) Orthogonally, so \(2\left(\frac{3-2 \lambda}{1+\lambda}\right)-4\left(\frac{2-3 \lambda}{1+\lambda}\right)=-12-11\) \(\Rightarrow 6-4 \lambda-8+12 \lambda=-23-23 \lambda\) \(\Rightarrow 31 \lambda=-21 \Rightarrow \lambda=-\frac{21}{31}\) \(\therefore\) Centre of the circle (i) is \(\left(-\frac{3+\frac{42}{31}}{1-\frac{21}{31}},-\frac{2+\frac{63}{31}}{1-\frac{21}{31}}\right)=\left(-\frac{135}{10},-\frac{125}{10}\right)=\left(-\frac{27}{2},-\frac{25}{2}\right)\)
TS EAMCET-11.09.2020
Conic Section
120064
If the radical axis of the circles \(x^2+y^2+2 \alpha x+2 \beta y+c=0\) and \(x^2+y^2+\frac{3}{2} x+4 y+c=0\) touches the circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}+2 \mathrm{y}+\mathbf{1}=\mathbf{0}\), then \(4 \alpha \beta-8 \alpha-3 \beta+10=\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Conic Section
120061
The radical centre of the system of circles, \(x^2+y^2+4 x+7=0,\) \(2\left(x^2+y^2\right)+3 x+5 y+9=0\) and \(x^2+y^2+y=0\) is
1 \((-2,-1)\)
2 \((1,-2)\)
3 \((-1,-2)\)
4 None of these
Explanation:
A We have:- The system of circles, \(x^2+y^2+4 x+7=0\) \(2\left(x^2+y^2\right)+3 x+5 y+9=0\) \(x^2+y^2+\frac{3}{2} x+\frac{5}{2} y+\frac{9}{2}=0\) \(x^2+y^2+y=0\) On subtracting \(\mathrm{Eq}^{\mathrm{n}}\) (ii) from \(\mathrm{Eq}^{\mathrm{n}}\) (i), we get \(4 x-\frac{3}{2} x-\frac{5}{2} y+7-\frac{9}{2}=0\) \(\frac{5}{2} x-\frac{5}{2} y+\frac{5}{2}=0\) \(x-y+1=0\) On subtracting \(\mathrm{Eq}^{\mathrm{n}}\) (iii) from \(\mathrm{Eq}^{\mathrm{n}}\) (i), we get \(\Rightarrow \quad\left(\mathrm{x}^2+\mathrm{y}^2+\mathrm{y}\right)-\left(\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}+7\right)=0\) \(4 \mathrm{x}-\mathrm{y}+7=0\) On solving Eq \({ }^n\) (iv) and (v), we get:- \(x=-2, \quad y=-1\)Hence, radical centre is \((-2,-1)\).
Manipal UGET-2020
Conic Section
120062
If the lengths of the tangents drawn from a point \(P\) to the three circles \(\mathrm{x}^2+\mathrm{y}^2-\mathbf{4}=\mathbf{0}\), \(x^2+y^2-2 x+3 y=0\) and \(x^2+y^2+7 y-18=0\) are equal, then the coordinates of \(P\) are
1 \((2,5)\)
2 \((3,4)\)
3 \((4,3)\)
4 \((5,2)\)
Explanation:
D Given equation of circle - \(x^2+y^2-4=0\) \(x^2+y^2-2 x+3 y=0\) \(\mathrm{x}^2+\mathrm{y}^2+7 \mathrm{y}-18=0\) Radical axis of the circle by equation (i) and (iii), we get- \(-7 y+14=0\) \(y=2\) By equation (i) and (ii) Radical axis of the circle \(2 x-3 y-4=0\) From equation (iv) and (v) \(2 \mathrm{x}-6-4=0\) \(\mathrm{x}=5\)Radical center \((5,2)=\mathrm{P}=(5,2)\)
AP EAMCET-08.07.2022
Conic Section
120063
The centre of the circle passing through the points of intersection of the circles \((x+3)^2+(y\) \(+2)^2=25\) and \((x-2)^2+(y-3)^2=25\) and cutting the circle \((x+1)^2+(y-2)^2=16\) orthogonally is
1 \(\left(\frac{-27}{2}, \frac{-25}{2}\right)\)
2 \((0,0)\)
3 \(\left(\frac{16}{3}, \frac{-25}{4}\right)\)
4 \(\left(\frac{4}{7}, \frac{3}{7}\right)\)
Explanation:
A Equation of circle passes through the point of intersections of given circles, \(\qquad \mathrm{S}_1 \equiv(\mathrm{x}+3)^2+(\mathrm{y}+2)^2=25\) \(\text { And, } \quad \mathrm{S}_2 \equiv(\mathrm{x}-2)^2+(\mathrm{y}-3)^2=25\) \(\text { is } \mathrm{S}_1+\lambda \mathrm{S}_2=0\) \((1+\lambda) \mathrm{x}^2+(1+\lambda) \mathrm{y}^2+(6-4 \lambda) \mathrm{x}+(4-6 \lambda) \mathrm{y}\) \(-12(1+\lambda)=0\) \(\mathrm{x}^2+\mathrm{y}^2+2\left(\frac{3-2 \lambda}{1+\lambda}\right) \mathrm{x}+2\left(\frac{2-3 \lambda}{1+\lambda}\right) \mathrm{y}-12=0\) And, Cuts the another given circle \((x+1)^2+(y-2)^2=16\) Orthogonally, so \(2\left(\frac{3-2 \lambda}{1+\lambda}\right)-4\left(\frac{2-3 \lambda}{1+\lambda}\right)=-12-11\) \(\Rightarrow 6-4 \lambda-8+12 \lambda=-23-23 \lambda\) \(\Rightarrow 31 \lambda=-21 \Rightarrow \lambda=-\frac{21}{31}\) \(\therefore\) Centre of the circle (i) is \(\left(-\frac{3+\frac{42}{31}}{1-\frac{21}{31}},-\frac{2+\frac{63}{31}}{1-\frac{21}{31}}\right)=\left(-\frac{135}{10},-\frac{125}{10}\right)=\left(-\frac{27}{2},-\frac{25}{2}\right)\)
TS EAMCET-11.09.2020
Conic Section
120064
If the radical axis of the circles \(x^2+y^2+2 \alpha x+2 \beta y+c=0\) and \(x^2+y^2+\frac{3}{2} x+4 y+c=0\) touches the circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}+2 \mathrm{y}+\mathbf{1}=\mathbf{0}\), then \(4 \alpha \beta-8 \alpha-3 \beta+10=\)