NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119672
If one of the diameters of the curve \(x^2+y^2-4 x\) \(-6 y+9=0\) is a chord of a circle with centre (1, 1), the radius of this circle is
1 3
2 2
3 \(\sqrt{2}\)
4 1
Explanation:
A Given equation of curve is \(x^2+y^2-4 x-6 y+9=0\) \((x-2)^2+(y-3)^2-4-9+9=0\) \((x-2)^2+(y-3)^2=4\) which is a circle with centre \((2,3)\) and radius 2 The diameter of this circle is a chord of a circle with a centre o \((1,1)\). \(\mathrm{OP}=\sqrt{(3-1)^2+(2-1)^2}\) \(\mathrm{OP}=\sqrt{5}\) \(\mathrm{QP}=2\) \(\therefore \quad \mathrm{r}^2=(\sqrt{5})^2+2^2\) \(r=3\)
WB JEE-2017
Conic Section
119689
If the lines \(3 x-4 y-7=0\) and \(2 x-3 y-5=0\) are two diameters of a circle of area \(49 \pi\), square units, the equation of the circle is
1 \(x^2+y^2+2 x-2 y-47=0\)
2 \(x^2+y^2-2 x+2 y-62=0\)
3 \(x^2+y^2+2 x-2 y-62=0\)
4 \(x^2+y^2-2 x+2 y-47=0\)
Explanation:
D Given equation- \(3 \mathrm{x}-4 \mathrm{y}-7=0\) \(2 \mathrm{x}-3 \mathrm{y}-5=0\) By equation (i) \(\times 2\) and (ii) \(\times 3\), we get - \(6 x-8 y-14=0\) \(6 x-9 y-15=0\) \(+\quad+\) \(y+1=0\) \(y=-1\) Put the value of \(y\) in the equation \(6 x-8 x(-1)-14=0\) \(6 x=6\) \(x=1\) since given area of the circle is \(\pi r^2=49 \pi\) \(r^2=49\) \(r=7\) For the centre \((1,-1)\) and radius \(\mathrm{r}=7\) Equation centre is - \((x-1)^2+(y+1)^2=(7)^2\) \(x^2+1-2 x+y^2+1+2 y=49\) \(x^2+y^2-2 x+2 y+2-49=0\) \(x^2+y^2-2 x+2 y-47=0\)
Assam CEE-2018
Conic Section
119673
Let \(P\left(a t^2, 2 a t\right), Q, R\left(a r^2, 2 a r\right)\) be three points on a parabola \(y^2=4 \mathrm{ax}\). If \(P Q\) is the focal chord and \(P K, Q R\) are parallel where the co-ordinates of \(K\) is \((2 a, 0)\), then the value of \(r\) is
1 \(\frac{t}{1-t^2}\)
2 \(\frac{1-t^2}{t}\)
3 \(\frac{t^2+1}{t}\)
4 \(\frac{t^2-1}{t}\)
Explanation:
D \(\text { Coordinates of } \mathrm{Q}=\left(\frac{\mathrm{a}}{\mathrm{t}^2}, \frac{-2 \mathrm{a}}{\mathrm{t}}\right)\) \(\mathrm{PK} \text { and } \mathrm{QC} \text { are parallel }\) \(\mathrm{M}_{\mathrm{pk}}=\mathrm{M}_{\mathrm{QR}}\) \(\frac{2 \mathrm{at}-0}{\mathrm{at}^2-2 \mathrm{a}}=\frac{2 \mathrm{ar}+\frac{2 \mathrm{a}}{\mathrm{t}}}{\mathrm{ar}^2-\frac{\mathrm{a}}{\mathrm{t}^2}}\) \(\frac{\mathrm{rt}}{\mathrm{t}^2-2}=\frac{2}{\mathrm{r}-\frac{1}{\mathrm{t}}}\) \(\mathrm{r}=\frac{\mathrm{t}^2-1}{\mathrm{t}}\)
WB JEE-2018
Conic Section
119674
The circle \(s=0\) cuts the circle \(x^2+y^2-4 x+2 y-7=0\) orthogonally. If \((2,3)\) is the centre of the circle \(S=0\), then its radius is
1 2
2 1
3 3
4 4
Explanation:
A : \(\text { Given angle } \mathrm{C}_1 \mathrm{OC}_2=90^{\circ}\) \(\text { As we know that, }\) \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{C}_1 \mathrm{C}_2\right)^2\) \(\mathrm{r}_1^2+12=16\) \(\mathrm{r}_1^2=4\) \(\mathrm{r}_1=2\)
119672
If one of the diameters of the curve \(x^2+y^2-4 x\) \(-6 y+9=0\) is a chord of a circle with centre (1, 1), the radius of this circle is
1 3
2 2
3 \(\sqrt{2}\)
4 1
Explanation:
A Given equation of curve is \(x^2+y^2-4 x-6 y+9=0\) \((x-2)^2+(y-3)^2-4-9+9=0\) \((x-2)^2+(y-3)^2=4\) which is a circle with centre \((2,3)\) and radius 2 The diameter of this circle is a chord of a circle with a centre o \((1,1)\). \(\mathrm{OP}=\sqrt{(3-1)^2+(2-1)^2}\) \(\mathrm{OP}=\sqrt{5}\) \(\mathrm{QP}=2\) \(\therefore \quad \mathrm{r}^2=(\sqrt{5})^2+2^2\) \(r=3\)
WB JEE-2017
Conic Section
119689
If the lines \(3 x-4 y-7=0\) and \(2 x-3 y-5=0\) are two diameters of a circle of area \(49 \pi\), square units, the equation of the circle is
1 \(x^2+y^2+2 x-2 y-47=0\)
2 \(x^2+y^2-2 x+2 y-62=0\)
3 \(x^2+y^2+2 x-2 y-62=0\)
4 \(x^2+y^2-2 x+2 y-47=0\)
Explanation:
D Given equation- \(3 \mathrm{x}-4 \mathrm{y}-7=0\) \(2 \mathrm{x}-3 \mathrm{y}-5=0\) By equation (i) \(\times 2\) and (ii) \(\times 3\), we get - \(6 x-8 y-14=0\) \(6 x-9 y-15=0\) \(+\quad+\) \(y+1=0\) \(y=-1\) Put the value of \(y\) in the equation \(6 x-8 x(-1)-14=0\) \(6 x=6\) \(x=1\) since given area of the circle is \(\pi r^2=49 \pi\) \(r^2=49\) \(r=7\) For the centre \((1,-1)\) and radius \(\mathrm{r}=7\) Equation centre is - \((x-1)^2+(y+1)^2=(7)^2\) \(x^2+1-2 x+y^2+1+2 y=49\) \(x^2+y^2-2 x+2 y+2-49=0\) \(x^2+y^2-2 x+2 y-47=0\)
Assam CEE-2018
Conic Section
119673
Let \(P\left(a t^2, 2 a t\right), Q, R\left(a r^2, 2 a r\right)\) be three points on a parabola \(y^2=4 \mathrm{ax}\). If \(P Q\) is the focal chord and \(P K, Q R\) are parallel where the co-ordinates of \(K\) is \((2 a, 0)\), then the value of \(r\) is
1 \(\frac{t}{1-t^2}\)
2 \(\frac{1-t^2}{t}\)
3 \(\frac{t^2+1}{t}\)
4 \(\frac{t^2-1}{t}\)
Explanation:
D \(\text { Coordinates of } \mathrm{Q}=\left(\frac{\mathrm{a}}{\mathrm{t}^2}, \frac{-2 \mathrm{a}}{\mathrm{t}}\right)\) \(\mathrm{PK} \text { and } \mathrm{QC} \text { are parallel }\) \(\mathrm{M}_{\mathrm{pk}}=\mathrm{M}_{\mathrm{QR}}\) \(\frac{2 \mathrm{at}-0}{\mathrm{at}^2-2 \mathrm{a}}=\frac{2 \mathrm{ar}+\frac{2 \mathrm{a}}{\mathrm{t}}}{\mathrm{ar}^2-\frac{\mathrm{a}}{\mathrm{t}^2}}\) \(\frac{\mathrm{rt}}{\mathrm{t}^2-2}=\frac{2}{\mathrm{r}-\frac{1}{\mathrm{t}}}\) \(\mathrm{r}=\frac{\mathrm{t}^2-1}{\mathrm{t}}\)
WB JEE-2018
Conic Section
119674
The circle \(s=0\) cuts the circle \(x^2+y^2-4 x+2 y-7=0\) orthogonally. If \((2,3)\) is the centre of the circle \(S=0\), then its radius is
1 2
2 1
3 3
4 4
Explanation:
A : \(\text { Given angle } \mathrm{C}_1 \mathrm{OC}_2=90^{\circ}\) \(\text { As we know that, }\) \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{C}_1 \mathrm{C}_2\right)^2\) \(\mathrm{r}_1^2+12=16\) \(\mathrm{r}_1^2=4\) \(\mathrm{r}_1=2\)
119672
If one of the diameters of the curve \(x^2+y^2-4 x\) \(-6 y+9=0\) is a chord of a circle with centre (1, 1), the radius of this circle is
1 3
2 2
3 \(\sqrt{2}\)
4 1
Explanation:
A Given equation of curve is \(x^2+y^2-4 x-6 y+9=0\) \((x-2)^2+(y-3)^2-4-9+9=0\) \((x-2)^2+(y-3)^2=4\) which is a circle with centre \((2,3)\) and radius 2 The diameter of this circle is a chord of a circle with a centre o \((1,1)\). \(\mathrm{OP}=\sqrt{(3-1)^2+(2-1)^2}\) \(\mathrm{OP}=\sqrt{5}\) \(\mathrm{QP}=2\) \(\therefore \quad \mathrm{r}^2=(\sqrt{5})^2+2^2\) \(r=3\)
WB JEE-2017
Conic Section
119689
If the lines \(3 x-4 y-7=0\) and \(2 x-3 y-5=0\) are two diameters of a circle of area \(49 \pi\), square units, the equation of the circle is
1 \(x^2+y^2+2 x-2 y-47=0\)
2 \(x^2+y^2-2 x+2 y-62=0\)
3 \(x^2+y^2+2 x-2 y-62=0\)
4 \(x^2+y^2-2 x+2 y-47=0\)
Explanation:
D Given equation- \(3 \mathrm{x}-4 \mathrm{y}-7=0\) \(2 \mathrm{x}-3 \mathrm{y}-5=0\) By equation (i) \(\times 2\) and (ii) \(\times 3\), we get - \(6 x-8 y-14=0\) \(6 x-9 y-15=0\) \(+\quad+\) \(y+1=0\) \(y=-1\) Put the value of \(y\) in the equation \(6 x-8 x(-1)-14=0\) \(6 x=6\) \(x=1\) since given area of the circle is \(\pi r^2=49 \pi\) \(r^2=49\) \(r=7\) For the centre \((1,-1)\) and radius \(\mathrm{r}=7\) Equation centre is - \((x-1)^2+(y+1)^2=(7)^2\) \(x^2+1-2 x+y^2+1+2 y=49\) \(x^2+y^2-2 x+2 y+2-49=0\) \(x^2+y^2-2 x+2 y-47=0\)
Assam CEE-2018
Conic Section
119673
Let \(P\left(a t^2, 2 a t\right), Q, R\left(a r^2, 2 a r\right)\) be three points on a parabola \(y^2=4 \mathrm{ax}\). If \(P Q\) is the focal chord and \(P K, Q R\) are parallel where the co-ordinates of \(K\) is \((2 a, 0)\), then the value of \(r\) is
1 \(\frac{t}{1-t^2}\)
2 \(\frac{1-t^2}{t}\)
3 \(\frac{t^2+1}{t}\)
4 \(\frac{t^2-1}{t}\)
Explanation:
D \(\text { Coordinates of } \mathrm{Q}=\left(\frac{\mathrm{a}}{\mathrm{t}^2}, \frac{-2 \mathrm{a}}{\mathrm{t}}\right)\) \(\mathrm{PK} \text { and } \mathrm{QC} \text { are parallel }\) \(\mathrm{M}_{\mathrm{pk}}=\mathrm{M}_{\mathrm{QR}}\) \(\frac{2 \mathrm{at}-0}{\mathrm{at}^2-2 \mathrm{a}}=\frac{2 \mathrm{ar}+\frac{2 \mathrm{a}}{\mathrm{t}}}{\mathrm{ar}^2-\frac{\mathrm{a}}{\mathrm{t}^2}}\) \(\frac{\mathrm{rt}}{\mathrm{t}^2-2}=\frac{2}{\mathrm{r}-\frac{1}{\mathrm{t}}}\) \(\mathrm{r}=\frac{\mathrm{t}^2-1}{\mathrm{t}}\)
WB JEE-2018
Conic Section
119674
The circle \(s=0\) cuts the circle \(x^2+y^2-4 x+2 y-7=0\) orthogonally. If \((2,3)\) is the centre of the circle \(S=0\), then its radius is
1 2
2 1
3 3
4 4
Explanation:
A : \(\text { Given angle } \mathrm{C}_1 \mathrm{OC}_2=90^{\circ}\) \(\text { As we know that, }\) \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{C}_1 \mathrm{C}_2\right)^2\) \(\mathrm{r}_1^2+12=16\) \(\mathrm{r}_1^2=4\) \(\mathrm{r}_1=2\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Conic Section
119672
If one of the diameters of the curve \(x^2+y^2-4 x\) \(-6 y+9=0\) is a chord of a circle with centre (1, 1), the radius of this circle is
1 3
2 2
3 \(\sqrt{2}\)
4 1
Explanation:
A Given equation of curve is \(x^2+y^2-4 x-6 y+9=0\) \((x-2)^2+(y-3)^2-4-9+9=0\) \((x-2)^2+(y-3)^2=4\) which is a circle with centre \((2,3)\) and radius 2 The diameter of this circle is a chord of a circle with a centre o \((1,1)\). \(\mathrm{OP}=\sqrt{(3-1)^2+(2-1)^2}\) \(\mathrm{OP}=\sqrt{5}\) \(\mathrm{QP}=2\) \(\therefore \quad \mathrm{r}^2=(\sqrt{5})^2+2^2\) \(r=3\)
WB JEE-2017
Conic Section
119689
If the lines \(3 x-4 y-7=0\) and \(2 x-3 y-5=0\) are two diameters of a circle of area \(49 \pi\), square units, the equation of the circle is
1 \(x^2+y^2+2 x-2 y-47=0\)
2 \(x^2+y^2-2 x+2 y-62=0\)
3 \(x^2+y^2+2 x-2 y-62=0\)
4 \(x^2+y^2-2 x+2 y-47=0\)
Explanation:
D Given equation- \(3 \mathrm{x}-4 \mathrm{y}-7=0\) \(2 \mathrm{x}-3 \mathrm{y}-5=0\) By equation (i) \(\times 2\) and (ii) \(\times 3\), we get - \(6 x-8 y-14=0\) \(6 x-9 y-15=0\) \(+\quad+\) \(y+1=0\) \(y=-1\) Put the value of \(y\) in the equation \(6 x-8 x(-1)-14=0\) \(6 x=6\) \(x=1\) since given area of the circle is \(\pi r^2=49 \pi\) \(r^2=49\) \(r=7\) For the centre \((1,-1)\) and radius \(\mathrm{r}=7\) Equation centre is - \((x-1)^2+(y+1)^2=(7)^2\) \(x^2+1-2 x+y^2+1+2 y=49\) \(x^2+y^2-2 x+2 y+2-49=0\) \(x^2+y^2-2 x+2 y-47=0\)
Assam CEE-2018
Conic Section
119673
Let \(P\left(a t^2, 2 a t\right), Q, R\left(a r^2, 2 a r\right)\) be three points on a parabola \(y^2=4 \mathrm{ax}\). If \(P Q\) is the focal chord and \(P K, Q R\) are parallel where the co-ordinates of \(K\) is \((2 a, 0)\), then the value of \(r\) is
1 \(\frac{t}{1-t^2}\)
2 \(\frac{1-t^2}{t}\)
3 \(\frac{t^2+1}{t}\)
4 \(\frac{t^2-1}{t}\)
Explanation:
D \(\text { Coordinates of } \mathrm{Q}=\left(\frac{\mathrm{a}}{\mathrm{t}^2}, \frac{-2 \mathrm{a}}{\mathrm{t}}\right)\) \(\mathrm{PK} \text { and } \mathrm{QC} \text { are parallel }\) \(\mathrm{M}_{\mathrm{pk}}=\mathrm{M}_{\mathrm{QR}}\) \(\frac{2 \mathrm{at}-0}{\mathrm{at}^2-2 \mathrm{a}}=\frac{2 \mathrm{ar}+\frac{2 \mathrm{a}}{\mathrm{t}}}{\mathrm{ar}^2-\frac{\mathrm{a}}{\mathrm{t}^2}}\) \(\frac{\mathrm{rt}}{\mathrm{t}^2-2}=\frac{2}{\mathrm{r}-\frac{1}{\mathrm{t}}}\) \(\mathrm{r}=\frac{\mathrm{t}^2-1}{\mathrm{t}}\)
WB JEE-2018
Conic Section
119674
The circle \(s=0\) cuts the circle \(x^2+y^2-4 x+2 y-7=0\) orthogonally. If \((2,3)\) is the centre of the circle \(S=0\), then its radius is
1 2
2 1
3 3
4 4
Explanation:
A : \(\text { Given angle } \mathrm{C}_1 \mathrm{OC}_2=90^{\circ}\) \(\text { As we know that, }\) \(\mathrm{r}_1^2+\mathrm{r}_2^2=\left(\mathrm{C}_1 \mathrm{C}_2\right)^2\) \(\mathrm{r}_1^2+12=16\) \(\mathrm{r}_1^2=4\) \(\mathrm{r}_1=2\)
