119752
For the circle \(x^2+y^2-2 x-4 y+3=0\), the point
1 \((0,1)\) lies on the circle Prove, \(x=0, \quad y=1\) \(0+1-0-4+3=0\) \(4-4=0 \quad \text { (lies on the circle) }\)
2 \((3,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(9+1-6-4+3=0\) \(3>0 \text { (lies outside the circle) }\)
3 \((1,3)\) lies inside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+9-2-12+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\)
4 \((1,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+1-2-4+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) \(-1\lt 0\) (lies inside the circle) So, a, b, c are correct.
Explanation:
A \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-4 \mathrm{y}+3=0\) (a.) \((0,1)\) lies on the circle Prove, \(x=0, \quad y=1\) \(0+1-0-4+3=0\) \(4-4=0 \quad \text { (lies on the circle) }\) (b.) \((3,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(9+1-6-4+3=0\) \(3>0 \text { (lies outside the circle) }\) (c.) \((1,3)\) lies inside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+9-2-12+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) (d.) \((1,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+1-2-4+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) \(-1\lt 0\) (lies inside the circle) So, a, b, c are correct.
UPSEE-2011
Conic Section
119753
The angle between the tangent drawn from origin to the circle \((x-7)^2+(y+1)^2=25\) is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{2}\)
4 \(\frac{\pi}{8}\)
Explanation:
C Let tangent from origin be \(\mathrm{y}=\mathrm{mx}\) Using the condition of tangency we get. \(\frac{7 \mathrm{~m}+1}{\sqrt{\mathrm{m}^2+1}}=5\) \((7 \mathrm{~m}+1)^2=\left(5 \sqrt{\mathrm{m}^2+1}\right)^2\) \(49 \mathrm{~m}^2+1+14 \mathrm{~m}=25\left(\mathrm{~m}^2+1\right)\) \(49 \mathrm{~m}^2-25 \mathrm{~m}^2+14 \mathrm{~m}-25+1=0\) \(24 \mathrm{~m}^2+14 \mathrm{~m}-24=0\) \(12 \mathrm{~m}^2+7 \mathrm{~m}-12=0\) \(12 \mathrm{~m}^2+16 \mathrm{~m}-9 \mathrm{~m}-12=0\) \(4 \mathrm{~m}(3 \mathrm{~m}+4)-3(3 \mathrm{~m}+4)=0\) \((3 \mathrm{~m}+4)(4 \mathrm{~m}-3)=0\) \(\mathrm{~m}=\frac{-4}{3}, \mathrm{~m}=\frac{3}{4} .\) Therefore the product of both the slopes is -1 , i.e. \(\frac{3}{4} \times \frac{-4}{3}=-1\)Hence the angle between the two tangent is \(\frac{\pi}{2}\).
UPSEE-2010
Conic Section
119754
\(A B C D\) is a square of unit area. \(A\) circle is tangent to two sides of \(\mathrm{ABCD}\) and passes through exactly one of its vertices. The radius of the circle is
1 \(2-\sqrt{2}\)
2 \(\sqrt{2}-1\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
A \(:\) Let the centre of the circle is \((\mathrm{r}, \mathrm{r})\) \(\mathrm{OE}=\mathrm{CE}=(1-\mathrm{r})\) Applying Pythagoras theorem in are \(\Delta \mathrm{OED}\) \((1-\mathrm{r})^2+(1-\mathrm{r})^2=\mathrm{r}^2\) \(2(1-\mathrm{r})^2=\mathrm{r}^2\) \(\sqrt{2}(1-\mathrm{r})=\mathrm{r}\) \(\mathrm{r}(\sqrt{2}+1)=\sqrt{2}\) \(\mathrm{r}=\frac{\sqrt{2}}{\sqrt{2}+1}=2-\sqrt{2}\)
AMU-2019
Conic Section
119755
If the line \(x+2 b y+7=0\) is a diameter of the circle \(x^2+y^2-6 x+2 y=0\), then \(b\) is equal to
1 -5
2 -3
3 2
4 5
Explanation:
D We have equation of circle is \(x^2+y^2-6 x+2 y=0\) Ans: (7) Exp: (7) : Given, that \(\mathrm{k}=\mathrm{r}\) \(\mathrm{h}=1\) \(\mathrm{OP}=\mathrm{r}, \mathrm{PR}=1\) \(\mathrm{OR}=\left|\frac{\mathrm{r}+1}{\sqrt{2}}\right|\) \(\mathrm{r}^2=1+\frac{(\mathrm{r}+1)^2}{2}\) \(2 \mathrm{r}^2=2+\mathrm{r}^2+1+2 \mathrm{r}\) \(\mathrm{r}^2-2 \mathrm{r}-3=0\) \((\mathrm{r}-3)(\mathrm{r}+1)=0\) \(\mathrm{r}=3,-1\) \(\mathrm{h}+\mathrm{k}+\mathrm{r}=1+3+3 \Rightarrow 7\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119752
For the circle \(x^2+y^2-2 x-4 y+3=0\), the point
1 \((0,1)\) lies on the circle Prove, \(x=0, \quad y=1\) \(0+1-0-4+3=0\) \(4-4=0 \quad \text { (lies on the circle) }\)
2 \((3,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(9+1-6-4+3=0\) \(3>0 \text { (lies outside the circle) }\)
3 \((1,3)\) lies inside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+9-2-12+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\)
4 \((1,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+1-2-4+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) \(-1\lt 0\) (lies inside the circle) So, a, b, c are correct.
Explanation:
A \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-4 \mathrm{y}+3=0\) (a.) \((0,1)\) lies on the circle Prove, \(x=0, \quad y=1\) \(0+1-0-4+3=0\) \(4-4=0 \quad \text { (lies on the circle) }\) (b.) \((3,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(9+1-6-4+3=0\) \(3>0 \text { (lies outside the circle) }\) (c.) \((1,3)\) lies inside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+9-2-12+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) (d.) \((1,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+1-2-4+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) \(-1\lt 0\) (lies inside the circle) So, a, b, c are correct.
UPSEE-2011
Conic Section
119753
The angle between the tangent drawn from origin to the circle \((x-7)^2+(y+1)^2=25\) is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{2}\)
4 \(\frac{\pi}{8}\)
Explanation:
C Let tangent from origin be \(\mathrm{y}=\mathrm{mx}\) Using the condition of tangency we get. \(\frac{7 \mathrm{~m}+1}{\sqrt{\mathrm{m}^2+1}}=5\) \((7 \mathrm{~m}+1)^2=\left(5 \sqrt{\mathrm{m}^2+1}\right)^2\) \(49 \mathrm{~m}^2+1+14 \mathrm{~m}=25\left(\mathrm{~m}^2+1\right)\) \(49 \mathrm{~m}^2-25 \mathrm{~m}^2+14 \mathrm{~m}-25+1=0\) \(24 \mathrm{~m}^2+14 \mathrm{~m}-24=0\) \(12 \mathrm{~m}^2+7 \mathrm{~m}-12=0\) \(12 \mathrm{~m}^2+16 \mathrm{~m}-9 \mathrm{~m}-12=0\) \(4 \mathrm{~m}(3 \mathrm{~m}+4)-3(3 \mathrm{~m}+4)=0\) \((3 \mathrm{~m}+4)(4 \mathrm{~m}-3)=0\) \(\mathrm{~m}=\frac{-4}{3}, \mathrm{~m}=\frac{3}{4} .\) Therefore the product of both the slopes is -1 , i.e. \(\frac{3}{4} \times \frac{-4}{3}=-1\)Hence the angle between the two tangent is \(\frac{\pi}{2}\).
UPSEE-2010
Conic Section
119754
\(A B C D\) is a square of unit area. \(A\) circle is tangent to two sides of \(\mathrm{ABCD}\) and passes through exactly one of its vertices. The radius of the circle is
1 \(2-\sqrt{2}\)
2 \(\sqrt{2}-1\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
A \(:\) Let the centre of the circle is \((\mathrm{r}, \mathrm{r})\) \(\mathrm{OE}=\mathrm{CE}=(1-\mathrm{r})\) Applying Pythagoras theorem in are \(\Delta \mathrm{OED}\) \((1-\mathrm{r})^2+(1-\mathrm{r})^2=\mathrm{r}^2\) \(2(1-\mathrm{r})^2=\mathrm{r}^2\) \(\sqrt{2}(1-\mathrm{r})=\mathrm{r}\) \(\mathrm{r}(\sqrt{2}+1)=\sqrt{2}\) \(\mathrm{r}=\frac{\sqrt{2}}{\sqrt{2}+1}=2-\sqrt{2}\)
AMU-2019
Conic Section
119755
If the line \(x+2 b y+7=0\) is a diameter of the circle \(x^2+y^2-6 x+2 y=0\), then \(b\) is equal to
1 -5
2 -3
3 2
4 5
Explanation:
D We have equation of circle is \(x^2+y^2-6 x+2 y=0\) Ans: (7) Exp: (7) : Given, that \(\mathrm{k}=\mathrm{r}\) \(\mathrm{h}=1\) \(\mathrm{OP}=\mathrm{r}, \mathrm{PR}=1\) \(\mathrm{OR}=\left|\frac{\mathrm{r}+1}{\sqrt{2}}\right|\) \(\mathrm{r}^2=1+\frac{(\mathrm{r}+1)^2}{2}\) \(2 \mathrm{r}^2=2+\mathrm{r}^2+1+2 \mathrm{r}\) \(\mathrm{r}^2-2 \mathrm{r}-3=0\) \((\mathrm{r}-3)(\mathrm{r}+1)=0\) \(\mathrm{r}=3,-1\) \(\mathrm{h}+\mathrm{k}+\mathrm{r}=1+3+3 \Rightarrow 7\)
119752
For the circle \(x^2+y^2-2 x-4 y+3=0\), the point
1 \((0,1)\) lies on the circle Prove, \(x=0, \quad y=1\) \(0+1-0-4+3=0\) \(4-4=0 \quad \text { (lies on the circle) }\)
2 \((3,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(9+1-6-4+3=0\) \(3>0 \text { (lies outside the circle) }\)
3 \((1,3)\) lies inside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+9-2-12+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\)
4 \((1,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+1-2-4+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) \(-1\lt 0\) (lies inside the circle) So, a, b, c are correct.
Explanation:
A \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-4 \mathrm{y}+3=0\) (a.) \((0,1)\) lies on the circle Prove, \(x=0, \quad y=1\) \(0+1-0-4+3=0\) \(4-4=0 \quad \text { (lies on the circle) }\) (b.) \((3,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(9+1-6-4+3=0\) \(3>0 \text { (lies outside the circle) }\) (c.) \((1,3)\) lies inside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+9-2-12+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) (d.) \((1,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+1-2-4+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) \(-1\lt 0\) (lies inside the circle) So, a, b, c are correct.
UPSEE-2011
Conic Section
119753
The angle between the tangent drawn from origin to the circle \((x-7)^2+(y+1)^2=25\) is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{2}\)
4 \(\frac{\pi}{8}\)
Explanation:
C Let tangent from origin be \(\mathrm{y}=\mathrm{mx}\) Using the condition of tangency we get. \(\frac{7 \mathrm{~m}+1}{\sqrt{\mathrm{m}^2+1}}=5\) \((7 \mathrm{~m}+1)^2=\left(5 \sqrt{\mathrm{m}^2+1}\right)^2\) \(49 \mathrm{~m}^2+1+14 \mathrm{~m}=25\left(\mathrm{~m}^2+1\right)\) \(49 \mathrm{~m}^2-25 \mathrm{~m}^2+14 \mathrm{~m}-25+1=0\) \(24 \mathrm{~m}^2+14 \mathrm{~m}-24=0\) \(12 \mathrm{~m}^2+7 \mathrm{~m}-12=0\) \(12 \mathrm{~m}^2+16 \mathrm{~m}-9 \mathrm{~m}-12=0\) \(4 \mathrm{~m}(3 \mathrm{~m}+4)-3(3 \mathrm{~m}+4)=0\) \((3 \mathrm{~m}+4)(4 \mathrm{~m}-3)=0\) \(\mathrm{~m}=\frac{-4}{3}, \mathrm{~m}=\frac{3}{4} .\) Therefore the product of both the slopes is -1 , i.e. \(\frac{3}{4} \times \frac{-4}{3}=-1\)Hence the angle between the two tangent is \(\frac{\pi}{2}\).
UPSEE-2010
Conic Section
119754
\(A B C D\) is a square of unit area. \(A\) circle is tangent to two sides of \(\mathrm{ABCD}\) and passes through exactly one of its vertices. The radius of the circle is
1 \(2-\sqrt{2}\)
2 \(\sqrt{2}-1\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
A \(:\) Let the centre of the circle is \((\mathrm{r}, \mathrm{r})\) \(\mathrm{OE}=\mathrm{CE}=(1-\mathrm{r})\) Applying Pythagoras theorem in are \(\Delta \mathrm{OED}\) \((1-\mathrm{r})^2+(1-\mathrm{r})^2=\mathrm{r}^2\) \(2(1-\mathrm{r})^2=\mathrm{r}^2\) \(\sqrt{2}(1-\mathrm{r})=\mathrm{r}\) \(\mathrm{r}(\sqrt{2}+1)=\sqrt{2}\) \(\mathrm{r}=\frac{\sqrt{2}}{\sqrt{2}+1}=2-\sqrt{2}\)
AMU-2019
Conic Section
119755
If the line \(x+2 b y+7=0\) is a diameter of the circle \(x^2+y^2-6 x+2 y=0\), then \(b\) is equal to
1 -5
2 -3
3 2
4 5
Explanation:
D We have equation of circle is \(x^2+y^2-6 x+2 y=0\) Ans: (7) Exp: (7) : Given, that \(\mathrm{k}=\mathrm{r}\) \(\mathrm{h}=1\) \(\mathrm{OP}=\mathrm{r}, \mathrm{PR}=1\) \(\mathrm{OR}=\left|\frac{\mathrm{r}+1}{\sqrt{2}}\right|\) \(\mathrm{r}^2=1+\frac{(\mathrm{r}+1)^2}{2}\) \(2 \mathrm{r}^2=2+\mathrm{r}^2+1+2 \mathrm{r}\) \(\mathrm{r}^2-2 \mathrm{r}-3=0\) \((\mathrm{r}-3)(\mathrm{r}+1)=0\) \(\mathrm{r}=3,-1\) \(\mathrm{h}+\mathrm{k}+\mathrm{r}=1+3+3 \Rightarrow 7\)
119752
For the circle \(x^2+y^2-2 x-4 y+3=0\), the point
1 \((0,1)\) lies on the circle Prove, \(x=0, \quad y=1\) \(0+1-0-4+3=0\) \(4-4=0 \quad \text { (lies on the circle) }\)
2 \((3,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(9+1-6-4+3=0\) \(3>0 \text { (lies outside the circle) }\)
3 \((1,3)\) lies inside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+9-2-12+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\)
4 \((1,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+1-2-4+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) \(-1\lt 0\) (lies inside the circle) So, a, b, c are correct.
Explanation:
A \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-4 \mathrm{y}+3=0\) (a.) \((0,1)\) lies on the circle Prove, \(x=0, \quad y=1\) \(0+1-0-4+3=0\) \(4-4=0 \quad \text { (lies on the circle) }\) (b.) \((3,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(9+1-6-4+3=0\) \(3>0 \text { (lies outside the circle) }\) (c.) \((1,3)\) lies inside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+9-2-12+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) (d.) \((1,1)\) lies outside the circle Prove, \(x^2+y^2-2 x-4 y+3=0\) \(1+1-2-4+3=0\) \(-1\lt 0 \text { (lies inside the circle) }\) \(-1\lt 0\) (lies inside the circle) So, a, b, c are correct.
UPSEE-2011
Conic Section
119753
The angle between the tangent drawn from origin to the circle \((x-7)^2+(y+1)^2=25\) is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{2}\)
4 \(\frac{\pi}{8}\)
Explanation:
C Let tangent from origin be \(\mathrm{y}=\mathrm{mx}\) Using the condition of tangency we get. \(\frac{7 \mathrm{~m}+1}{\sqrt{\mathrm{m}^2+1}}=5\) \((7 \mathrm{~m}+1)^2=\left(5 \sqrt{\mathrm{m}^2+1}\right)^2\) \(49 \mathrm{~m}^2+1+14 \mathrm{~m}=25\left(\mathrm{~m}^2+1\right)\) \(49 \mathrm{~m}^2-25 \mathrm{~m}^2+14 \mathrm{~m}-25+1=0\) \(24 \mathrm{~m}^2+14 \mathrm{~m}-24=0\) \(12 \mathrm{~m}^2+7 \mathrm{~m}-12=0\) \(12 \mathrm{~m}^2+16 \mathrm{~m}-9 \mathrm{~m}-12=0\) \(4 \mathrm{~m}(3 \mathrm{~m}+4)-3(3 \mathrm{~m}+4)=0\) \((3 \mathrm{~m}+4)(4 \mathrm{~m}-3)=0\) \(\mathrm{~m}=\frac{-4}{3}, \mathrm{~m}=\frac{3}{4} .\) Therefore the product of both the slopes is -1 , i.e. \(\frac{3}{4} \times \frac{-4}{3}=-1\)Hence the angle between the two tangent is \(\frac{\pi}{2}\).
UPSEE-2010
Conic Section
119754
\(A B C D\) is a square of unit area. \(A\) circle is tangent to two sides of \(\mathrm{ABCD}\) and passes through exactly one of its vertices. The radius of the circle is
1 \(2-\sqrt{2}\)
2 \(\sqrt{2}-1\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{\sqrt{2}}\)
Explanation:
A \(:\) Let the centre of the circle is \((\mathrm{r}, \mathrm{r})\) \(\mathrm{OE}=\mathrm{CE}=(1-\mathrm{r})\) Applying Pythagoras theorem in are \(\Delta \mathrm{OED}\) \((1-\mathrm{r})^2+(1-\mathrm{r})^2=\mathrm{r}^2\) \(2(1-\mathrm{r})^2=\mathrm{r}^2\) \(\sqrt{2}(1-\mathrm{r})=\mathrm{r}\) \(\mathrm{r}(\sqrt{2}+1)=\sqrt{2}\) \(\mathrm{r}=\frac{\sqrt{2}}{\sqrt{2}+1}=2-\sqrt{2}\)
AMU-2019
Conic Section
119755
If the line \(x+2 b y+7=0\) is a diameter of the circle \(x^2+y^2-6 x+2 y=0\), then \(b\) is equal to
1 -5
2 -3
3 2
4 5
Explanation:
D We have equation of circle is \(x^2+y^2-6 x+2 y=0\) Ans: (7) Exp: (7) : Given, that \(\mathrm{k}=\mathrm{r}\) \(\mathrm{h}=1\) \(\mathrm{OP}=\mathrm{r}, \mathrm{PR}=1\) \(\mathrm{OR}=\left|\frac{\mathrm{r}+1}{\sqrt{2}}\right|\) \(\mathrm{r}^2=1+\frac{(\mathrm{r}+1)^2}{2}\) \(2 \mathrm{r}^2=2+\mathrm{r}^2+1+2 \mathrm{r}\) \(\mathrm{r}^2-2 \mathrm{r}-3=0\) \((\mathrm{r}-3)(\mathrm{r}+1)=0\) \(\mathrm{r}=3,-1\) \(\mathrm{h}+\mathrm{k}+\mathrm{r}=1+3+3 \Rightarrow 7\)
