119738
If \(\theta\) is a parameter, then the parametric equations of the circle \(x^2+y^2-6 x+4 y-3=0\) are given by
1 \(x=3+4 \sin \theta\) and \(y=2+4 \cos \theta\)
2 \(x=3+4 \cos \theta\) and \(y=-2+4 \sin \theta\)
3 \(x=-3+4 \sin \theta\) and \(y=-2+4 \cos \theta\)
4 \(\mathrm{x}=3+4 \cos \theta\) and \(\mathrm{y}=2+4 \sin \theta\)
Explanation:
B Given that, Equation of circle is \(x^2+y^2-6 x+4 y-3=0\) \(\therefore \quad\left(\mathrm{x}^2-6 \mathrm{x}+9\right)-9+\left(\mathrm{y}^2+4 \mathrm{y}+4\right)-4-3=0\) \(\therefore \quad(\mathrm{x}-3)^2+(\mathrm{y}+2)^2=16\) \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}+\mathrm{k})^2=\mathrm{r}^2\) \(\text { From equation (i) and (ii) }\) \(\text { we get }\) \(\mathrm{h}=3, \mathrm{k}=-2, \mathrm{r}=4\) \(\mathrm{x}=\mathrm{h}+\mathrm{r} \cos \theta\) \(\mathrm{x}=3+4 \cos \theta\) \(y=-2+4 \sin \theta\) Parametric from is \(\mathrm{x}=\mathrm{h}+\mathrm{r} \cos \theta\) \(\mathrm{x}=3+4 \cos \theta\)And, \(\mathrm{y}=\mathrm{k}+\mathrm{r} \sin \theta\)
MHT CET-2020
Conic Section
119739
The Cartesian equation of the curve given by \(x=6 \cos \theta, y=6 \sin \theta\) is
1 \(x^2+y^2=6\)
2 \(x^2+y^2=5\)
3 \(x^2+y^2=16\)
4 \(x^2+y^2=36\)
Explanation:
D Giventhat, \(\mathrm{x}=6 \cos \theta\) \(\mathrm{y}=6 \sin \theta\) \(x^2+y^2=36 \cos ^2 \theta+36 \sin ^2 \theta\) \(=36\left(\cos ^2 \theta+\sin ^2\right)\) \(\mathrm{x}^2+\mathrm{y}^2=36\) Squares on equation (i) and (ii) and add it, \(\mathrm{x}^2+\mathrm{y}^2 =36 \cos ^2 \theta+36 \sin ^2 \theta\) \(=36\left(\cos ^2 \theta+\sin ^2\right)\) \(\mathrm{x}^2+\mathrm{y}^2 =36\)
MHT CET-2020
Conic Section
119741
If \(\mathbf{A}(3,-2,2), \mathbf{B}(2, \lambda+1,5)\) are the end points of the diameter of the circle and if the point \((5,6,-1)\) lies on the circle, then \(\lambda=\)
1 8
2 7
3 6
4 5
Explanation:
A Given that, The point are \((3,-2,2)\) and The angle subtended at the point \(P\) in the semicircle \(\mathrm{APB}\) is a right angle. \(\mathrm{AP} \perp \mathrm{PB}\), we get, Now direction ratio of AP are \((5-3,6+2,-1-2)\) \((2,8,-3)\) \(\text { Now direction ratio of PB }\) \((2-5, \lambda+1-6,5+1)\) \((-3, \lambda-5,6)\) \(\text { Since AP } \perp \mathrm{PB}, \text { we write }\) \((2)(-3)+8(\lambda-5)-3(6)=0\) \(-6+8 \lambda-40-18=0\) \(\lambda=8\)
MHT CET-2020
Conic Section
119742
The centre and radius of a circle \(\mathbf{x}=\mathbf{a}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{8 a t}{1+t^2}\), are respectively
1 \((0,0)\) and 2a units
2 \((0,0)\) and \(4 \mathrm{a}\) units
3 \((0,0)\) and \(3 a\) units
4 \((0,0)\) and \(3 \mathrm{a}\) units
Explanation:
B Given equation of circle in parametric form is \(\mathrm{x}=4 \mathrm{a}\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)\) \(y=4 a\left(\frac{2 t}{1+t^2}\right)\) \(\mathrm{x}=4 \mathrm{a} \cos 2 \theta\) \(\mathrm{y}=4 \mathrm{a} \sin 2 \theta, \quad \text { where } \mathrm{t}=\tan \theta\) \(y=r \sin \theta\) \(r=4 a \text {, centre }(0,0)\) Comparing with \(x=r \cos \theta\)
MHT CET-2020
Conic Section
119743
If \((a, b)\) and \((4,3)\) are end-points of a diameter of the circle \(x^2+y^2+4 x-6 y+11=0\), then \((a, b)=\)
1 \((-8,3)\)
2 \((8,3)\)
3 \((8,-3)\)
4 \((-8,-3)\)
Explanation:
A Given that, Equation of circle, \(x^2+y^2+4 x-6 y+11=0\) \((a, b)\) and \((4,3)\) are end points of a diameter of the \(\text { circle } x^2+y^2+4 x-6 y+11=0\) \(\therefore\) equation of circle is \(x^2-y^2+(-4-a) x +(-3-b) y\) \(+(4 a+3 b)=0\) Comparing with equation (i) and (ii) we get \(-4-a=4\) \(-3-b=-6 \Rightarrow a=-8 \text { and } b=3\)
119738
If \(\theta\) is a parameter, then the parametric equations of the circle \(x^2+y^2-6 x+4 y-3=0\) are given by
1 \(x=3+4 \sin \theta\) and \(y=2+4 \cos \theta\)
2 \(x=3+4 \cos \theta\) and \(y=-2+4 \sin \theta\)
3 \(x=-3+4 \sin \theta\) and \(y=-2+4 \cos \theta\)
4 \(\mathrm{x}=3+4 \cos \theta\) and \(\mathrm{y}=2+4 \sin \theta\)
Explanation:
B Given that, Equation of circle is \(x^2+y^2-6 x+4 y-3=0\) \(\therefore \quad\left(\mathrm{x}^2-6 \mathrm{x}+9\right)-9+\left(\mathrm{y}^2+4 \mathrm{y}+4\right)-4-3=0\) \(\therefore \quad(\mathrm{x}-3)^2+(\mathrm{y}+2)^2=16\) \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}+\mathrm{k})^2=\mathrm{r}^2\) \(\text { From equation (i) and (ii) }\) \(\text { we get }\) \(\mathrm{h}=3, \mathrm{k}=-2, \mathrm{r}=4\) \(\mathrm{x}=\mathrm{h}+\mathrm{r} \cos \theta\) \(\mathrm{x}=3+4 \cos \theta\) \(y=-2+4 \sin \theta\) Parametric from is \(\mathrm{x}=\mathrm{h}+\mathrm{r} \cos \theta\) \(\mathrm{x}=3+4 \cos \theta\)And, \(\mathrm{y}=\mathrm{k}+\mathrm{r} \sin \theta\)
MHT CET-2020
Conic Section
119739
The Cartesian equation of the curve given by \(x=6 \cos \theta, y=6 \sin \theta\) is
1 \(x^2+y^2=6\)
2 \(x^2+y^2=5\)
3 \(x^2+y^2=16\)
4 \(x^2+y^2=36\)
Explanation:
D Giventhat, \(\mathrm{x}=6 \cos \theta\) \(\mathrm{y}=6 \sin \theta\) \(x^2+y^2=36 \cos ^2 \theta+36 \sin ^2 \theta\) \(=36\left(\cos ^2 \theta+\sin ^2\right)\) \(\mathrm{x}^2+\mathrm{y}^2=36\) Squares on equation (i) and (ii) and add it, \(\mathrm{x}^2+\mathrm{y}^2 =36 \cos ^2 \theta+36 \sin ^2 \theta\) \(=36\left(\cos ^2 \theta+\sin ^2\right)\) \(\mathrm{x}^2+\mathrm{y}^2 =36\)
MHT CET-2020
Conic Section
119741
If \(\mathbf{A}(3,-2,2), \mathbf{B}(2, \lambda+1,5)\) are the end points of the diameter of the circle and if the point \((5,6,-1)\) lies on the circle, then \(\lambda=\)
1 8
2 7
3 6
4 5
Explanation:
A Given that, The point are \((3,-2,2)\) and The angle subtended at the point \(P\) in the semicircle \(\mathrm{APB}\) is a right angle. \(\mathrm{AP} \perp \mathrm{PB}\), we get, Now direction ratio of AP are \((5-3,6+2,-1-2)\) \((2,8,-3)\) \(\text { Now direction ratio of PB }\) \((2-5, \lambda+1-6,5+1)\) \((-3, \lambda-5,6)\) \(\text { Since AP } \perp \mathrm{PB}, \text { we write }\) \((2)(-3)+8(\lambda-5)-3(6)=0\) \(-6+8 \lambda-40-18=0\) \(\lambda=8\)
MHT CET-2020
Conic Section
119742
The centre and radius of a circle \(\mathbf{x}=\mathbf{a}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{8 a t}{1+t^2}\), are respectively
1 \((0,0)\) and 2a units
2 \((0,0)\) and \(4 \mathrm{a}\) units
3 \((0,0)\) and \(3 a\) units
4 \((0,0)\) and \(3 \mathrm{a}\) units
Explanation:
B Given equation of circle in parametric form is \(\mathrm{x}=4 \mathrm{a}\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)\) \(y=4 a\left(\frac{2 t}{1+t^2}\right)\) \(\mathrm{x}=4 \mathrm{a} \cos 2 \theta\) \(\mathrm{y}=4 \mathrm{a} \sin 2 \theta, \quad \text { where } \mathrm{t}=\tan \theta\) \(y=r \sin \theta\) \(r=4 a \text {, centre }(0,0)\) Comparing with \(x=r \cos \theta\)
MHT CET-2020
Conic Section
119743
If \((a, b)\) and \((4,3)\) are end-points of a diameter of the circle \(x^2+y^2+4 x-6 y+11=0\), then \((a, b)=\)
1 \((-8,3)\)
2 \((8,3)\)
3 \((8,-3)\)
4 \((-8,-3)\)
Explanation:
A Given that, Equation of circle, \(x^2+y^2+4 x-6 y+11=0\) \((a, b)\) and \((4,3)\) are end points of a diameter of the \(\text { circle } x^2+y^2+4 x-6 y+11=0\) \(\therefore\) equation of circle is \(x^2-y^2+(-4-a) x +(-3-b) y\) \(+(4 a+3 b)=0\) Comparing with equation (i) and (ii) we get \(-4-a=4\) \(-3-b=-6 \Rightarrow a=-8 \text { and } b=3\)
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Conic Section
119738
If \(\theta\) is a parameter, then the parametric equations of the circle \(x^2+y^2-6 x+4 y-3=0\) are given by
1 \(x=3+4 \sin \theta\) and \(y=2+4 \cos \theta\)
2 \(x=3+4 \cos \theta\) and \(y=-2+4 \sin \theta\)
3 \(x=-3+4 \sin \theta\) and \(y=-2+4 \cos \theta\)
4 \(\mathrm{x}=3+4 \cos \theta\) and \(\mathrm{y}=2+4 \sin \theta\)
Explanation:
B Given that, Equation of circle is \(x^2+y^2-6 x+4 y-3=0\) \(\therefore \quad\left(\mathrm{x}^2-6 \mathrm{x}+9\right)-9+\left(\mathrm{y}^2+4 \mathrm{y}+4\right)-4-3=0\) \(\therefore \quad(\mathrm{x}-3)^2+(\mathrm{y}+2)^2=16\) \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}+\mathrm{k})^2=\mathrm{r}^2\) \(\text { From equation (i) and (ii) }\) \(\text { we get }\) \(\mathrm{h}=3, \mathrm{k}=-2, \mathrm{r}=4\) \(\mathrm{x}=\mathrm{h}+\mathrm{r} \cos \theta\) \(\mathrm{x}=3+4 \cos \theta\) \(y=-2+4 \sin \theta\) Parametric from is \(\mathrm{x}=\mathrm{h}+\mathrm{r} \cos \theta\) \(\mathrm{x}=3+4 \cos \theta\)And, \(\mathrm{y}=\mathrm{k}+\mathrm{r} \sin \theta\)
MHT CET-2020
Conic Section
119739
The Cartesian equation of the curve given by \(x=6 \cos \theta, y=6 \sin \theta\) is
1 \(x^2+y^2=6\)
2 \(x^2+y^2=5\)
3 \(x^2+y^2=16\)
4 \(x^2+y^2=36\)
Explanation:
D Giventhat, \(\mathrm{x}=6 \cos \theta\) \(\mathrm{y}=6 \sin \theta\) \(x^2+y^2=36 \cos ^2 \theta+36 \sin ^2 \theta\) \(=36\left(\cos ^2 \theta+\sin ^2\right)\) \(\mathrm{x}^2+\mathrm{y}^2=36\) Squares on equation (i) and (ii) and add it, \(\mathrm{x}^2+\mathrm{y}^2 =36 \cos ^2 \theta+36 \sin ^2 \theta\) \(=36\left(\cos ^2 \theta+\sin ^2\right)\) \(\mathrm{x}^2+\mathrm{y}^2 =36\)
MHT CET-2020
Conic Section
119741
If \(\mathbf{A}(3,-2,2), \mathbf{B}(2, \lambda+1,5)\) are the end points of the diameter of the circle and if the point \((5,6,-1)\) lies on the circle, then \(\lambda=\)
1 8
2 7
3 6
4 5
Explanation:
A Given that, The point are \((3,-2,2)\) and The angle subtended at the point \(P\) in the semicircle \(\mathrm{APB}\) is a right angle. \(\mathrm{AP} \perp \mathrm{PB}\), we get, Now direction ratio of AP are \((5-3,6+2,-1-2)\) \((2,8,-3)\) \(\text { Now direction ratio of PB }\) \((2-5, \lambda+1-6,5+1)\) \((-3, \lambda-5,6)\) \(\text { Since AP } \perp \mathrm{PB}, \text { we write }\) \((2)(-3)+8(\lambda-5)-3(6)=0\) \(-6+8 \lambda-40-18=0\) \(\lambda=8\)
MHT CET-2020
Conic Section
119742
The centre and radius of a circle \(\mathbf{x}=\mathbf{a}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{8 a t}{1+t^2}\), are respectively
1 \((0,0)\) and 2a units
2 \((0,0)\) and \(4 \mathrm{a}\) units
3 \((0,0)\) and \(3 a\) units
4 \((0,0)\) and \(3 \mathrm{a}\) units
Explanation:
B Given equation of circle in parametric form is \(\mathrm{x}=4 \mathrm{a}\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)\) \(y=4 a\left(\frac{2 t}{1+t^2}\right)\) \(\mathrm{x}=4 \mathrm{a} \cos 2 \theta\) \(\mathrm{y}=4 \mathrm{a} \sin 2 \theta, \quad \text { where } \mathrm{t}=\tan \theta\) \(y=r \sin \theta\) \(r=4 a \text {, centre }(0,0)\) Comparing with \(x=r \cos \theta\)
MHT CET-2020
Conic Section
119743
If \((a, b)\) and \((4,3)\) are end-points of a diameter of the circle \(x^2+y^2+4 x-6 y+11=0\), then \((a, b)=\)
1 \((-8,3)\)
2 \((8,3)\)
3 \((8,-3)\)
4 \((-8,-3)\)
Explanation:
A Given that, Equation of circle, \(x^2+y^2+4 x-6 y+11=0\) \((a, b)\) and \((4,3)\) are end points of a diameter of the \(\text { circle } x^2+y^2+4 x-6 y+11=0\) \(\therefore\) equation of circle is \(x^2-y^2+(-4-a) x +(-3-b) y\) \(+(4 a+3 b)=0\) Comparing with equation (i) and (ii) we get \(-4-a=4\) \(-3-b=-6 \Rightarrow a=-8 \text { and } b=3\)
119738
If \(\theta\) is a parameter, then the parametric equations of the circle \(x^2+y^2-6 x+4 y-3=0\) are given by
1 \(x=3+4 \sin \theta\) and \(y=2+4 \cos \theta\)
2 \(x=3+4 \cos \theta\) and \(y=-2+4 \sin \theta\)
3 \(x=-3+4 \sin \theta\) and \(y=-2+4 \cos \theta\)
4 \(\mathrm{x}=3+4 \cos \theta\) and \(\mathrm{y}=2+4 \sin \theta\)
Explanation:
B Given that, Equation of circle is \(x^2+y^2-6 x+4 y-3=0\) \(\therefore \quad\left(\mathrm{x}^2-6 \mathrm{x}+9\right)-9+\left(\mathrm{y}^2+4 \mathrm{y}+4\right)-4-3=0\) \(\therefore \quad(\mathrm{x}-3)^2+(\mathrm{y}+2)^2=16\) \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}+\mathrm{k})^2=\mathrm{r}^2\) \(\text { From equation (i) and (ii) }\) \(\text { we get }\) \(\mathrm{h}=3, \mathrm{k}=-2, \mathrm{r}=4\) \(\mathrm{x}=\mathrm{h}+\mathrm{r} \cos \theta\) \(\mathrm{x}=3+4 \cos \theta\) \(y=-2+4 \sin \theta\) Parametric from is \(\mathrm{x}=\mathrm{h}+\mathrm{r} \cos \theta\) \(\mathrm{x}=3+4 \cos \theta\)And, \(\mathrm{y}=\mathrm{k}+\mathrm{r} \sin \theta\)
MHT CET-2020
Conic Section
119739
The Cartesian equation of the curve given by \(x=6 \cos \theta, y=6 \sin \theta\) is
1 \(x^2+y^2=6\)
2 \(x^2+y^2=5\)
3 \(x^2+y^2=16\)
4 \(x^2+y^2=36\)
Explanation:
D Giventhat, \(\mathrm{x}=6 \cos \theta\) \(\mathrm{y}=6 \sin \theta\) \(x^2+y^2=36 \cos ^2 \theta+36 \sin ^2 \theta\) \(=36\left(\cos ^2 \theta+\sin ^2\right)\) \(\mathrm{x}^2+\mathrm{y}^2=36\) Squares on equation (i) and (ii) and add it, \(\mathrm{x}^2+\mathrm{y}^2 =36 \cos ^2 \theta+36 \sin ^2 \theta\) \(=36\left(\cos ^2 \theta+\sin ^2\right)\) \(\mathrm{x}^2+\mathrm{y}^2 =36\)
MHT CET-2020
Conic Section
119741
If \(\mathbf{A}(3,-2,2), \mathbf{B}(2, \lambda+1,5)\) are the end points of the diameter of the circle and if the point \((5,6,-1)\) lies on the circle, then \(\lambda=\)
1 8
2 7
3 6
4 5
Explanation:
A Given that, The point are \((3,-2,2)\) and The angle subtended at the point \(P\) in the semicircle \(\mathrm{APB}\) is a right angle. \(\mathrm{AP} \perp \mathrm{PB}\), we get, Now direction ratio of AP are \((5-3,6+2,-1-2)\) \((2,8,-3)\) \(\text { Now direction ratio of PB }\) \((2-5, \lambda+1-6,5+1)\) \((-3, \lambda-5,6)\) \(\text { Since AP } \perp \mathrm{PB}, \text { we write }\) \((2)(-3)+8(\lambda-5)-3(6)=0\) \(-6+8 \lambda-40-18=0\) \(\lambda=8\)
MHT CET-2020
Conic Section
119742
The centre and radius of a circle \(\mathbf{x}=\mathbf{a}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{8 a t}{1+t^2}\), are respectively
1 \((0,0)\) and 2a units
2 \((0,0)\) and \(4 \mathrm{a}\) units
3 \((0,0)\) and \(3 a\) units
4 \((0,0)\) and \(3 \mathrm{a}\) units
Explanation:
B Given equation of circle in parametric form is \(\mathrm{x}=4 \mathrm{a}\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)\) \(y=4 a\left(\frac{2 t}{1+t^2}\right)\) \(\mathrm{x}=4 \mathrm{a} \cos 2 \theta\) \(\mathrm{y}=4 \mathrm{a} \sin 2 \theta, \quad \text { where } \mathrm{t}=\tan \theta\) \(y=r \sin \theta\) \(r=4 a \text {, centre }(0,0)\) Comparing with \(x=r \cos \theta\)
MHT CET-2020
Conic Section
119743
If \((a, b)\) and \((4,3)\) are end-points of a diameter of the circle \(x^2+y^2+4 x-6 y+11=0\), then \((a, b)=\)
1 \((-8,3)\)
2 \((8,3)\)
3 \((8,-3)\)
4 \((-8,-3)\)
Explanation:
A Given that, Equation of circle, \(x^2+y^2+4 x-6 y+11=0\) \((a, b)\) and \((4,3)\) are end points of a diameter of the \(\text { circle } x^2+y^2+4 x-6 y+11=0\) \(\therefore\) equation of circle is \(x^2-y^2+(-4-a) x +(-3-b) y\) \(+(4 a+3 b)=0\) Comparing with equation (i) and (ii) we get \(-4-a=4\) \(-3-b=-6 \Rightarrow a=-8 \text { and } b=3\)
119738
If \(\theta\) is a parameter, then the parametric equations of the circle \(x^2+y^2-6 x+4 y-3=0\) are given by
1 \(x=3+4 \sin \theta\) and \(y=2+4 \cos \theta\)
2 \(x=3+4 \cos \theta\) and \(y=-2+4 \sin \theta\)
3 \(x=-3+4 \sin \theta\) and \(y=-2+4 \cos \theta\)
4 \(\mathrm{x}=3+4 \cos \theta\) and \(\mathrm{y}=2+4 \sin \theta\)
Explanation:
B Given that, Equation of circle is \(x^2+y^2-6 x+4 y-3=0\) \(\therefore \quad\left(\mathrm{x}^2-6 \mathrm{x}+9\right)-9+\left(\mathrm{y}^2+4 \mathrm{y}+4\right)-4-3=0\) \(\therefore \quad(\mathrm{x}-3)^2+(\mathrm{y}+2)^2=16\) \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}+\mathrm{k})^2=\mathrm{r}^2\) \(\text { From equation (i) and (ii) }\) \(\text { we get }\) \(\mathrm{h}=3, \mathrm{k}=-2, \mathrm{r}=4\) \(\mathrm{x}=\mathrm{h}+\mathrm{r} \cos \theta\) \(\mathrm{x}=3+4 \cos \theta\) \(y=-2+4 \sin \theta\) Parametric from is \(\mathrm{x}=\mathrm{h}+\mathrm{r} \cos \theta\) \(\mathrm{x}=3+4 \cos \theta\)And, \(\mathrm{y}=\mathrm{k}+\mathrm{r} \sin \theta\)
MHT CET-2020
Conic Section
119739
The Cartesian equation of the curve given by \(x=6 \cos \theta, y=6 \sin \theta\) is
1 \(x^2+y^2=6\)
2 \(x^2+y^2=5\)
3 \(x^2+y^2=16\)
4 \(x^2+y^2=36\)
Explanation:
D Giventhat, \(\mathrm{x}=6 \cos \theta\) \(\mathrm{y}=6 \sin \theta\) \(x^2+y^2=36 \cos ^2 \theta+36 \sin ^2 \theta\) \(=36\left(\cos ^2 \theta+\sin ^2\right)\) \(\mathrm{x}^2+\mathrm{y}^2=36\) Squares on equation (i) and (ii) and add it, \(\mathrm{x}^2+\mathrm{y}^2 =36 \cos ^2 \theta+36 \sin ^2 \theta\) \(=36\left(\cos ^2 \theta+\sin ^2\right)\) \(\mathrm{x}^2+\mathrm{y}^2 =36\)
MHT CET-2020
Conic Section
119741
If \(\mathbf{A}(3,-2,2), \mathbf{B}(2, \lambda+1,5)\) are the end points of the diameter of the circle and if the point \((5,6,-1)\) lies on the circle, then \(\lambda=\)
1 8
2 7
3 6
4 5
Explanation:
A Given that, The point are \((3,-2,2)\) and The angle subtended at the point \(P\) in the semicircle \(\mathrm{APB}\) is a right angle. \(\mathrm{AP} \perp \mathrm{PB}\), we get, Now direction ratio of AP are \((5-3,6+2,-1-2)\) \((2,8,-3)\) \(\text { Now direction ratio of PB }\) \((2-5, \lambda+1-6,5+1)\) \((-3, \lambda-5,6)\) \(\text { Since AP } \perp \mathrm{PB}, \text { we write }\) \((2)(-3)+8(\lambda-5)-3(6)=0\) \(-6+8 \lambda-40-18=0\) \(\lambda=8\)
MHT CET-2020
Conic Section
119742
The centre and radius of a circle \(\mathbf{x}=\mathbf{a}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{8 a t}{1+t^2}\), are respectively
1 \((0,0)\) and 2a units
2 \((0,0)\) and \(4 \mathrm{a}\) units
3 \((0,0)\) and \(3 a\) units
4 \((0,0)\) and \(3 \mathrm{a}\) units
Explanation:
B Given equation of circle in parametric form is \(\mathrm{x}=4 \mathrm{a}\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)\) \(y=4 a\left(\frac{2 t}{1+t^2}\right)\) \(\mathrm{x}=4 \mathrm{a} \cos 2 \theta\) \(\mathrm{y}=4 \mathrm{a} \sin 2 \theta, \quad \text { where } \mathrm{t}=\tan \theta\) \(y=r \sin \theta\) \(r=4 a \text {, centre }(0,0)\) Comparing with \(x=r \cos \theta\)
MHT CET-2020
Conic Section
119743
If \((a, b)\) and \((4,3)\) are end-points of a diameter of the circle \(x^2+y^2+4 x-6 y+11=0\), then \((a, b)=\)
1 \((-8,3)\)
2 \((8,3)\)
3 \((8,-3)\)
4 \((-8,-3)\)
Explanation:
A Given that, Equation of circle, \(x^2+y^2+4 x-6 y+11=0\) \((a, b)\) and \((4,3)\) are end points of a diameter of the \(\text { circle } x^2+y^2+4 x-6 y+11=0\) \(\therefore\) equation of circle is \(x^2-y^2+(-4-a) x +(-3-b) y\) \(+(4 a+3 b)=0\) Comparing with equation (i) and (ii) we get \(-4-a=4\) \(-3-b=-6 \Rightarrow a=-8 \text { and } b=3\)
