119683
Let \(\mathrm{C}\) be the circle with centre at \((1,1)\) and radius 1 . If \(T\) is the circle centred at \((0, k)\) passing through origin and touching the circle \(C\) externally, then the radius of \(T\) is equal to
1 \(\frac{\sqrt{3}}{\sqrt{2}}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{4}\)
Explanation:
D Given, centre \( (1,1)\) \(\text { radius } \mathrm{r}=1\) \(\text { distance between their centers- }\) \(\qquad \mathrm{k}+1=\sqrt{1+(\mathrm{k}-1)^2}\) \(\mathrm{k}+1=\sqrt{1+\mathrm{k}^2+1-2 \mathrm{k}}\) \(\text { On squaring both sides }\) \(\mathrm{k}+1)^2=1+\mathrm{k}^2+1-2 \mathrm{k}\) \(\mathrm{k}+1+2 \mathrm{k}=\mathrm{k}^2+2-2 \mathrm{k}\) \(\mathrm{k}=2-1\) \(\mathrm{k}=\frac{1}{4}\) On squaring both sides \((\mathrm{k}+1)^2=1+\mathrm{k}^2+1-2 \mathrm{k}\) \(\mathrm{k}^2+1+2 \mathrm{k}=\mathrm{k}^2+2-2 \mathrm{k}\) \(4 \mathrm{k}=2-1\) \(\mathrm{k}=\frac{1}{4}\)
JEE Main-2014
Conic Section
119684
Consider a circle \(\mathrm{C}\) which touches the \(\mathrm{Y}\)-axis at \((0,6)\) and cuts off an intercept \(6 \sqrt{5}\) on the \(X\) axis. Then the radius of the circle \(C\) is equal to
1 \(\sqrt{53}\)
2 9
3 8
4 \(\sqrt{82}\)
Explanation:
B Given point \((0,6)\) and intercept \(6 \sqrt{5}\) Then, From figure, \(\mathrm{r}^2=6^2+(3 \sqrt{5})^3\) \(\mathrm{r}^2=36+45\) \(\mathrm{r}^2=81\) \(\mathrm{r}=9\)
JEE Main 27.07.2021
Conic Section
119685
Let \(P\) and \(Q\) be two distinct points on a circle which has centre at \(C(2,3)\) and which passes through origin 0 . If \(O C\) is perpendicular to both the line segments \(C P\) and \(C Q\), then the set \(\{\mathbf{P}, \mathbf{Q}\}\) is equal to
D Given point \((2,3)\) Equation of circle- \(\mathrm{r}=\sqrt{2^2+3^2}=\sqrt{13}\) \((\mathrm{x}-2)^2+(\mathrm{y}-3)^2=13\) \(\mathrm{~m}_1=\frac{3}{2}, \quad \mathrm{~m}_2=\frac{-2}{3}=\tan \theta\) \(\sin \theta=\frac{2}{\sqrt{13}}, \cos \theta=\frac{-3}{\sqrt{13}}\) Coordinates of P.Q \((2 \pm \sqrt{13} \cos \theta, 3 \pm \sqrt{13} \sin \theta)\) \(\left(2 \pm \sqrt{13}\left(\frac{-3}{\sqrt{13}}\right), 3 \pm \sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right)\) \((2 \pm(-3), 3 \pm 2)\) \(\{(-1,5),(5,1)\}\)
JEE Main 27.07.2021
Conic Section
119686
A circle passes through the centre of another circle \(x^2+y^2-3 x-4 y-1=0\) and whose centre is \((5,2)\). Then the equation of this circle is..... #[Qdiff: Hard, QCat: Numerical Based, examname:
A Given circle equation - \(x^2+y^2-3 x-4 y-1=0\) Ans: c Exp: (C) : Given circle equation, \(x^2+y^2-4 x-2 y-20=0\) \(g=-2, \quad f=-1 \quad c=-20\) \(r=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+(-1)^2+20}=5\) at point \((10,7)\) \(\mathrm{CP}=\sqrt{(10-2)^2+(7-1)^2}=\sqrt{64+36}=10\) Thus the greatest or maximum distance of point from the given circle is - \(\mathrm{D}=\mathrm{CP}+\mathrm{r}, \quad \mathrm{D}=10+5, \quad \mathrm{D}=15\)
(b.) 10
Conic Section
119687
If the equation of tangent to a circle at point \((3,5)\) is \(2 x-y-1=0\) and its centre lies on \(x+y\) \(=5\), then the equation of circle is
1 \(x^2+y^2+6 x-16 y+28=0\)
2 \(x^2+y^2-6 x-16 y+28=0\)
3 \(x^2+y^2+6 x+6 y+28=0\)
4 \(x^2+y^2-6 x-6 y-28=0\)
Explanation:
A Given, equation of tangent \(2 \mathrm{x}-\mathrm{y}-1=0\) \(\mathrm{y}=2 \mathrm{x}-1\) \(\mathrm{~m}_1=2\) \(\mathrm{~m}_1 \mathrm{~m}_2=-1\) \(2 \times \mathrm{m}_2=-1\) \(\mathrm{~m}_2=-\frac{1}{2}\) Equation of perpendicular line choice passes through \((3,5) \text { and slope }=-\frac{1}{2}\) \(y-5=\frac{-1}{2}(x-3)\) \(2 y-10=-x+3\) \(2 y+x=13\) \(x+y=5 \text { (given) }\) By equation (i) \& (ii) \(x+2 y-13=0\) \(\pm x \pm y \mp 5=0\) \(y=8\) \(x=-3\) Radius \((\mathrm{r})=\sqrt{(3+3)^2+(5-8)^2}=\sqrt{36+9}=\sqrt{45}\) Equation of circle with centre \((-3,8) \&\) radius \(=3 \sqrt{5}\) \((x+3)^2+(y-8)^2=(3 \sqrt{5})^2\) \(x^2+9+6 x+y^2+64-16 y=45\) \(x^2+y^2+6 x-16 y+64+9-45=0\) \(x^2+y^2+6 x-16 y+28=0\)
119683
Let \(\mathrm{C}\) be the circle with centre at \((1,1)\) and radius 1 . If \(T\) is the circle centred at \((0, k)\) passing through origin and touching the circle \(C\) externally, then the radius of \(T\) is equal to
1 \(\frac{\sqrt{3}}{\sqrt{2}}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{4}\)
Explanation:
D Given, centre \( (1,1)\) \(\text { radius } \mathrm{r}=1\) \(\text { distance between their centers- }\) \(\qquad \mathrm{k}+1=\sqrt{1+(\mathrm{k}-1)^2}\) \(\mathrm{k}+1=\sqrt{1+\mathrm{k}^2+1-2 \mathrm{k}}\) \(\text { On squaring both sides }\) \(\mathrm{k}+1)^2=1+\mathrm{k}^2+1-2 \mathrm{k}\) \(\mathrm{k}+1+2 \mathrm{k}=\mathrm{k}^2+2-2 \mathrm{k}\) \(\mathrm{k}=2-1\) \(\mathrm{k}=\frac{1}{4}\) On squaring both sides \((\mathrm{k}+1)^2=1+\mathrm{k}^2+1-2 \mathrm{k}\) \(\mathrm{k}^2+1+2 \mathrm{k}=\mathrm{k}^2+2-2 \mathrm{k}\) \(4 \mathrm{k}=2-1\) \(\mathrm{k}=\frac{1}{4}\)
JEE Main-2014
Conic Section
119684
Consider a circle \(\mathrm{C}\) which touches the \(\mathrm{Y}\)-axis at \((0,6)\) and cuts off an intercept \(6 \sqrt{5}\) on the \(X\) axis. Then the radius of the circle \(C\) is equal to
1 \(\sqrt{53}\)
2 9
3 8
4 \(\sqrt{82}\)
Explanation:
B Given point \((0,6)\) and intercept \(6 \sqrt{5}\) Then, From figure, \(\mathrm{r}^2=6^2+(3 \sqrt{5})^3\) \(\mathrm{r}^2=36+45\) \(\mathrm{r}^2=81\) \(\mathrm{r}=9\)
JEE Main 27.07.2021
Conic Section
119685
Let \(P\) and \(Q\) be two distinct points on a circle which has centre at \(C(2,3)\) and which passes through origin 0 . If \(O C\) is perpendicular to both the line segments \(C P\) and \(C Q\), then the set \(\{\mathbf{P}, \mathbf{Q}\}\) is equal to
D Given point \((2,3)\) Equation of circle- \(\mathrm{r}=\sqrt{2^2+3^2}=\sqrt{13}\) \((\mathrm{x}-2)^2+(\mathrm{y}-3)^2=13\) \(\mathrm{~m}_1=\frac{3}{2}, \quad \mathrm{~m}_2=\frac{-2}{3}=\tan \theta\) \(\sin \theta=\frac{2}{\sqrt{13}}, \cos \theta=\frac{-3}{\sqrt{13}}\) Coordinates of P.Q \((2 \pm \sqrt{13} \cos \theta, 3 \pm \sqrt{13} \sin \theta)\) \(\left(2 \pm \sqrt{13}\left(\frac{-3}{\sqrt{13}}\right), 3 \pm \sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right)\) \((2 \pm(-3), 3 \pm 2)\) \(\{(-1,5),(5,1)\}\)
JEE Main 27.07.2021
Conic Section
119686
A circle passes through the centre of another circle \(x^2+y^2-3 x-4 y-1=0\) and whose centre is \((5,2)\). Then the equation of this circle is..... #[Qdiff: Hard, QCat: Numerical Based, examname:
A Given circle equation - \(x^2+y^2-3 x-4 y-1=0\) Ans: c Exp: (C) : Given circle equation, \(x^2+y^2-4 x-2 y-20=0\) \(g=-2, \quad f=-1 \quad c=-20\) \(r=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+(-1)^2+20}=5\) at point \((10,7)\) \(\mathrm{CP}=\sqrt{(10-2)^2+(7-1)^2}=\sqrt{64+36}=10\) Thus the greatest or maximum distance of point from the given circle is - \(\mathrm{D}=\mathrm{CP}+\mathrm{r}, \quad \mathrm{D}=10+5, \quad \mathrm{D}=15\)
(b.) 10
Conic Section
119687
If the equation of tangent to a circle at point \((3,5)\) is \(2 x-y-1=0\) and its centre lies on \(x+y\) \(=5\), then the equation of circle is
1 \(x^2+y^2+6 x-16 y+28=0\)
2 \(x^2+y^2-6 x-16 y+28=0\)
3 \(x^2+y^2+6 x+6 y+28=0\)
4 \(x^2+y^2-6 x-6 y-28=0\)
Explanation:
A Given, equation of tangent \(2 \mathrm{x}-\mathrm{y}-1=0\) \(\mathrm{y}=2 \mathrm{x}-1\) \(\mathrm{~m}_1=2\) \(\mathrm{~m}_1 \mathrm{~m}_2=-1\) \(2 \times \mathrm{m}_2=-1\) \(\mathrm{~m}_2=-\frac{1}{2}\) Equation of perpendicular line choice passes through \((3,5) \text { and slope }=-\frac{1}{2}\) \(y-5=\frac{-1}{2}(x-3)\) \(2 y-10=-x+3\) \(2 y+x=13\) \(x+y=5 \text { (given) }\) By equation (i) \& (ii) \(x+2 y-13=0\) \(\pm x \pm y \mp 5=0\) \(y=8\) \(x=-3\) Radius \((\mathrm{r})=\sqrt{(3+3)^2+(5-8)^2}=\sqrt{36+9}=\sqrt{45}\) Equation of circle with centre \((-3,8) \&\) radius \(=3 \sqrt{5}\) \((x+3)^2+(y-8)^2=(3 \sqrt{5})^2\) \(x^2+9+6 x+y^2+64-16 y=45\) \(x^2+y^2+6 x-16 y+64+9-45=0\) \(x^2+y^2+6 x-16 y+28=0\)
119683
Let \(\mathrm{C}\) be the circle with centre at \((1,1)\) and radius 1 . If \(T\) is the circle centred at \((0, k)\) passing through origin and touching the circle \(C\) externally, then the radius of \(T\) is equal to
1 \(\frac{\sqrt{3}}{\sqrt{2}}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{4}\)
Explanation:
D Given, centre \( (1,1)\) \(\text { radius } \mathrm{r}=1\) \(\text { distance between their centers- }\) \(\qquad \mathrm{k}+1=\sqrt{1+(\mathrm{k}-1)^2}\) \(\mathrm{k}+1=\sqrt{1+\mathrm{k}^2+1-2 \mathrm{k}}\) \(\text { On squaring both sides }\) \(\mathrm{k}+1)^2=1+\mathrm{k}^2+1-2 \mathrm{k}\) \(\mathrm{k}+1+2 \mathrm{k}=\mathrm{k}^2+2-2 \mathrm{k}\) \(\mathrm{k}=2-1\) \(\mathrm{k}=\frac{1}{4}\) On squaring both sides \((\mathrm{k}+1)^2=1+\mathrm{k}^2+1-2 \mathrm{k}\) \(\mathrm{k}^2+1+2 \mathrm{k}=\mathrm{k}^2+2-2 \mathrm{k}\) \(4 \mathrm{k}=2-1\) \(\mathrm{k}=\frac{1}{4}\)
JEE Main-2014
Conic Section
119684
Consider a circle \(\mathrm{C}\) which touches the \(\mathrm{Y}\)-axis at \((0,6)\) and cuts off an intercept \(6 \sqrt{5}\) on the \(X\) axis. Then the radius of the circle \(C\) is equal to
1 \(\sqrt{53}\)
2 9
3 8
4 \(\sqrt{82}\)
Explanation:
B Given point \((0,6)\) and intercept \(6 \sqrt{5}\) Then, From figure, \(\mathrm{r}^2=6^2+(3 \sqrt{5})^3\) \(\mathrm{r}^2=36+45\) \(\mathrm{r}^2=81\) \(\mathrm{r}=9\)
JEE Main 27.07.2021
Conic Section
119685
Let \(P\) and \(Q\) be two distinct points on a circle which has centre at \(C(2,3)\) and which passes through origin 0 . If \(O C\) is perpendicular to both the line segments \(C P\) and \(C Q\), then the set \(\{\mathbf{P}, \mathbf{Q}\}\) is equal to
D Given point \((2,3)\) Equation of circle- \(\mathrm{r}=\sqrt{2^2+3^2}=\sqrt{13}\) \((\mathrm{x}-2)^2+(\mathrm{y}-3)^2=13\) \(\mathrm{~m}_1=\frac{3}{2}, \quad \mathrm{~m}_2=\frac{-2}{3}=\tan \theta\) \(\sin \theta=\frac{2}{\sqrt{13}}, \cos \theta=\frac{-3}{\sqrt{13}}\) Coordinates of P.Q \((2 \pm \sqrt{13} \cos \theta, 3 \pm \sqrt{13} \sin \theta)\) \(\left(2 \pm \sqrt{13}\left(\frac{-3}{\sqrt{13}}\right), 3 \pm \sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right)\) \((2 \pm(-3), 3 \pm 2)\) \(\{(-1,5),(5,1)\}\)
JEE Main 27.07.2021
Conic Section
119686
A circle passes through the centre of another circle \(x^2+y^2-3 x-4 y-1=0\) and whose centre is \((5,2)\). Then the equation of this circle is..... #[Qdiff: Hard, QCat: Numerical Based, examname:
A Given circle equation - \(x^2+y^2-3 x-4 y-1=0\) Ans: c Exp: (C) : Given circle equation, \(x^2+y^2-4 x-2 y-20=0\) \(g=-2, \quad f=-1 \quad c=-20\) \(r=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+(-1)^2+20}=5\) at point \((10,7)\) \(\mathrm{CP}=\sqrt{(10-2)^2+(7-1)^2}=\sqrt{64+36}=10\) Thus the greatest or maximum distance of point from the given circle is - \(\mathrm{D}=\mathrm{CP}+\mathrm{r}, \quad \mathrm{D}=10+5, \quad \mathrm{D}=15\)
(b.) 10
Conic Section
119687
If the equation of tangent to a circle at point \((3,5)\) is \(2 x-y-1=0\) and its centre lies on \(x+y\) \(=5\), then the equation of circle is
1 \(x^2+y^2+6 x-16 y+28=0\)
2 \(x^2+y^2-6 x-16 y+28=0\)
3 \(x^2+y^2+6 x+6 y+28=0\)
4 \(x^2+y^2-6 x-6 y-28=0\)
Explanation:
A Given, equation of tangent \(2 \mathrm{x}-\mathrm{y}-1=0\) \(\mathrm{y}=2 \mathrm{x}-1\) \(\mathrm{~m}_1=2\) \(\mathrm{~m}_1 \mathrm{~m}_2=-1\) \(2 \times \mathrm{m}_2=-1\) \(\mathrm{~m}_2=-\frac{1}{2}\) Equation of perpendicular line choice passes through \((3,5) \text { and slope }=-\frac{1}{2}\) \(y-5=\frac{-1}{2}(x-3)\) \(2 y-10=-x+3\) \(2 y+x=13\) \(x+y=5 \text { (given) }\) By equation (i) \& (ii) \(x+2 y-13=0\) \(\pm x \pm y \mp 5=0\) \(y=8\) \(x=-3\) Radius \((\mathrm{r})=\sqrt{(3+3)^2+(5-8)^2}=\sqrt{36+9}=\sqrt{45}\) Equation of circle with centre \((-3,8) \&\) radius \(=3 \sqrt{5}\) \((x+3)^2+(y-8)^2=(3 \sqrt{5})^2\) \(x^2+9+6 x+y^2+64-16 y=45\) \(x^2+y^2+6 x-16 y+64+9-45=0\) \(x^2+y^2+6 x-16 y+28=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119683
Let \(\mathrm{C}\) be the circle with centre at \((1,1)\) and radius 1 . If \(T\) is the circle centred at \((0, k)\) passing through origin and touching the circle \(C\) externally, then the radius of \(T\) is equal to
1 \(\frac{\sqrt{3}}{\sqrt{2}}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{4}\)
Explanation:
D Given, centre \( (1,1)\) \(\text { radius } \mathrm{r}=1\) \(\text { distance between their centers- }\) \(\qquad \mathrm{k}+1=\sqrt{1+(\mathrm{k}-1)^2}\) \(\mathrm{k}+1=\sqrt{1+\mathrm{k}^2+1-2 \mathrm{k}}\) \(\text { On squaring both sides }\) \(\mathrm{k}+1)^2=1+\mathrm{k}^2+1-2 \mathrm{k}\) \(\mathrm{k}+1+2 \mathrm{k}=\mathrm{k}^2+2-2 \mathrm{k}\) \(\mathrm{k}=2-1\) \(\mathrm{k}=\frac{1}{4}\) On squaring both sides \((\mathrm{k}+1)^2=1+\mathrm{k}^2+1-2 \mathrm{k}\) \(\mathrm{k}^2+1+2 \mathrm{k}=\mathrm{k}^2+2-2 \mathrm{k}\) \(4 \mathrm{k}=2-1\) \(\mathrm{k}=\frac{1}{4}\)
JEE Main-2014
Conic Section
119684
Consider a circle \(\mathrm{C}\) which touches the \(\mathrm{Y}\)-axis at \((0,6)\) and cuts off an intercept \(6 \sqrt{5}\) on the \(X\) axis. Then the radius of the circle \(C\) is equal to
1 \(\sqrt{53}\)
2 9
3 8
4 \(\sqrt{82}\)
Explanation:
B Given point \((0,6)\) and intercept \(6 \sqrt{5}\) Then, From figure, \(\mathrm{r}^2=6^2+(3 \sqrt{5})^3\) \(\mathrm{r}^2=36+45\) \(\mathrm{r}^2=81\) \(\mathrm{r}=9\)
JEE Main 27.07.2021
Conic Section
119685
Let \(P\) and \(Q\) be two distinct points on a circle which has centre at \(C(2,3)\) and which passes through origin 0 . If \(O C\) is perpendicular to both the line segments \(C P\) and \(C Q\), then the set \(\{\mathbf{P}, \mathbf{Q}\}\) is equal to
D Given point \((2,3)\) Equation of circle- \(\mathrm{r}=\sqrt{2^2+3^2}=\sqrt{13}\) \((\mathrm{x}-2)^2+(\mathrm{y}-3)^2=13\) \(\mathrm{~m}_1=\frac{3}{2}, \quad \mathrm{~m}_2=\frac{-2}{3}=\tan \theta\) \(\sin \theta=\frac{2}{\sqrt{13}}, \cos \theta=\frac{-3}{\sqrt{13}}\) Coordinates of P.Q \((2 \pm \sqrt{13} \cos \theta, 3 \pm \sqrt{13} \sin \theta)\) \(\left(2 \pm \sqrt{13}\left(\frac{-3}{\sqrt{13}}\right), 3 \pm \sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right)\) \((2 \pm(-3), 3 \pm 2)\) \(\{(-1,5),(5,1)\}\)
JEE Main 27.07.2021
Conic Section
119686
A circle passes through the centre of another circle \(x^2+y^2-3 x-4 y-1=0\) and whose centre is \((5,2)\). Then the equation of this circle is..... #[Qdiff: Hard, QCat: Numerical Based, examname:
A Given circle equation - \(x^2+y^2-3 x-4 y-1=0\) Ans: c Exp: (C) : Given circle equation, \(x^2+y^2-4 x-2 y-20=0\) \(g=-2, \quad f=-1 \quad c=-20\) \(r=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+(-1)^2+20}=5\) at point \((10,7)\) \(\mathrm{CP}=\sqrt{(10-2)^2+(7-1)^2}=\sqrt{64+36}=10\) Thus the greatest or maximum distance of point from the given circle is - \(\mathrm{D}=\mathrm{CP}+\mathrm{r}, \quad \mathrm{D}=10+5, \quad \mathrm{D}=15\)
(b.) 10
Conic Section
119687
If the equation of tangent to a circle at point \((3,5)\) is \(2 x-y-1=0\) and its centre lies on \(x+y\) \(=5\), then the equation of circle is
1 \(x^2+y^2+6 x-16 y+28=0\)
2 \(x^2+y^2-6 x-16 y+28=0\)
3 \(x^2+y^2+6 x+6 y+28=0\)
4 \(x^2+y^2-6 x-6 y-28=0\)
Explanation:
A Given, equation of tangent \(2 \mathrm{x}-\mathrm{y}-1=0\) \(\mathrm{y}=2 \mathrm{x}-1\) \(\mathrm{~m}_1=2\) \(\mathrm{~m}_1 \mathrm{~m}_2=-1\) \(2 \times \mathrm{m}_2=-1\) \(\mathrm{~m}_2=-\frac{1}{2}\) Equation of perpendicular line choice passes through \((3,5) \text { and slope }=-\frac{1}{2}\) \(y-5=\frac{-1}{2}(x-3)\) \(2 y-10=-x+3\) \(2 y+x=13\) \(x+y=5 \text { (given) }\) By equation (i) \& (ii) \(x+2 y-13=0\) \(\pm x \pm y \mp 5=0\) \(y=8\) \(x=-3\) Radius \((\mathrm{r})=\sqrt{(3+3)^2+(5-8)^2}=\sqrt{36+9}=\sqrt{45}\) Equation of circle with centre \((-3,8) \&\) radius \(=3 \sqrt{5}\) \((x+3)^2+(y-8)^2=(3 \sqrt{5})^2\) \(x^2+9+6 x+y^2+64-16 y=45\) \(x^2+y^2+6 x-16 y+64+9-45=0\) \(x^2+y^2+6 x-16 y+28=0\)
119683
Let \(\mathrm{C}\) be the circle with centre at \((1,1)\) and radius 1 . If \(T\) is the circle centred at \((0, k)\) passing through origin and touching the circle \(C\) externally, then the radius of \(T\) is equal to
1 \(\frac{\sqrt{3}}{\sqrt{2}}\)
2 \(\frac{\sqrt{3}}{2}\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{4}\)
Explanation:
D Given, centre \( (1,1)\) \(\text { radius } \mathrm{r}=1\) \(\text { distance between their centers- }\) \(\qquad \mathrm{k}+1=\sqrt{1+(\mathrm{k}-1)^2}\) \(\mathrm{k}+1=\sqrt{1+\mathrm{k}^2+1-2 \mathrm{k}}\) \(\text { On squaring both sides }\) \(\mathrm{k}+1)^2=1+\mathrm{k}^2+1-2 \mathrm{k}\) \(\mathrm{k}+1+2 \mathrm{k}=\mathrm{k}^2+2-2 \mathrm{k}\) \(\mathrm{k}=2-1\) \(\mathrm{k}=\frac{1}{4}\) On squaring both sides \((\mathrm{k}+1)^2=1+\mathrm{k}^2+1-2 \mathrm{k}\) \(\mathrm{k}^2+1+2 \mathrm{k}=\mathrm{k}^2+2-2 \mathrm{k}\) \(4 \mathrm{k}=2-1\) \(\mathrm{k}=\frac{1}{4}\)
JEE Main-2014
Conic Section
119684
Consider a circle \(\mathrm{C}\) which touches the \(\mathrm{Y}\)-axis at \((0,6)\) and cuts off an intercept \(6 \sqrt{5}\) on the \(X\) axis. Then the radius of the circle \(C\) is equal to
1 \(\sqrt{53}\)
2 9
3 8
4 \(\sqrt{82}\)
Explanation:
B Given point \((0,6)\) and intercept \(6 \sqrt{5}\) Then, From figure, \(\mathrm{r}^2=6^2+(3 \sqrt{5})^3\) \(\mathrm{r}^2=36+45\) \(\mathrm{r}^2=81\) \(\mathrm{r}=9\)
JEE Main 27.07.2021
Conic Section
119685
Let \(P\) and \(Q\) be two distinct points on a circle which has centre at \(C(2,3)\) and which passes through origin 0 . If \(O C\) is perpendicular to both the line segments \(C P\) and \(C Q\), then the set \(\{\mathbf{P}, \mathbf{Q}\}\) is equal to
D Given point \((2,3)\) Equation of circle- \(\mathrm{r}=\sqrt{2^2+3^2}=\sqrt{13}\) \((\mathrm{x}-2)^2+(\mathrm{y}-3)^2=13\) \(\mathrm{~m}_1=\frac{3}{2}, \quad \mathrm{~m}_2=\frac{-2}{3}=\tan \theta\) \(\sin \theta=\frac{2}{\sqrt{13}}, \cos \theta=\frac{-3}{\sqrt{13}}\) Coordinates of P.Q \((2 \pm \sqrt{13} \cos \theta, 3 \pm \sqrt{13} \sin \theta)\) \(\left(2 \pm \sqrt{13}\left(\frac{-3}{\sqrt{13}}\right), 3 \pm \sqrt{13}\left(\frac{2}{\sqrt{13}}\right)\right)\) \((2 \pm(-3), 3 \pm 2)\) \(\{(-1,5),(5,1)\}\)
JEE Main 27.07.2021
Conic Section
119686
A circle passes through the centre of another circle \(x^2+y^2-3 x-4 y-1=0\) and whose centre is \((5,2)\). Then the equation of this circle is..... #[Qdiff: Hard, QCat: Numerical Based, examname:
A Given circle equation - \(x^2+y^2-3 x-4 y-1=0\) Ans: c Exp: (C) : Given circle equation, \(x^2+y^2-4 x-2 y-20=0\) \(g=-2, \quad f=-1 \quad c=-20\) \(r=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+(-1)^2+20}=5\) at point \((10,7)\) \(\mathrm{CP}=\sqrt{(10-2)^2+(7-1)^2}=\sqrt{64+36}=10\) Thus the greatest or maximum distance of point from the given circle is - \(\mathrm{D}=\mathrm{CP}+\mathrm{r}, \quad \mathrm{D}=10+5, \quad \mathrm{D}=15\)
(b.) 10
Conic Section
119687
If the equation of tangent to a circle at point \((3,5)\) is \(2 x-y-1=0\) and its centre lies on \(x+y\) \(=5\), then the equation of circle is
1 \(x^2+y^2+6 x-16 y+28=0\)
2 \(x^2+y^2-6 x-16 y+28=0\)
3 \(x^2+y^2+6 x+6 y+28=0\)
4 \(x^2+y^2-6 x-6 y-28=0\)
Explanation:
A Given, equation of tangent \(2 \mathrm{x}-\mathrm{y}-1=0\) \(\mathrm{y}=2 \mathrm{x}-1\) \(\mathrm{~m}_1=2\) \(\mathrm{~m}_1 \mathrm{~m}_2=-1\) \(2 \times \mathrm{m}_2=-1\) \(\mathrm{~m}_2=-\frac{1}{2}\) Equation of perpendicular line choice passes through \((3,5) \text { and slope }=-\frac{1}{2}\) \(y-5=\frac{-1}{2}(x-3)\) \(2 y-10=-x+3\) \(2 y+x=13\) \(x+y=5 \text { (given) }\) By equation (i) \& (ii) \(x+2 y-13=0\) \(\pm x \pm y \mp 5=0\) \(y=8\) \(x=-3\) Radius \((\mathrm{r})=\sqrt{(3+3)^2+(5-8)^2}=\sqrt{36+9}=\sqrt{45}\) Equation of circle with centre \((-3,8) \&\) radius \(=3 \sqrt{5}\) \((x+3)^2+(y-8)^2=(3 \sqrt{5})^2\) \(x^2+9+6 x+y^2+64-16 y=45\) \(x^2+y^2+6 x-16 y+64+9-45=0\) \(x^2+y^2+6 x-16 y+28=0\)
