119667
The equations of the lines through \((1,1)\) and making angles of \(45^{\circ}\) with the line \(x+y=0\) are
1 \(x-1=0, x-y=0\)
2 \(x-y=0, y-1=0\)
3 \(x+y-2=0, y-1=0\)
4 \(x-1=0, y-1=0\)
Explanation:
D Given, equation of line \(\mathrm{x}+\mathrm{y}=0\) and \(\theta=45^{\circ}\) \(y=-x\) Slope of line \(\mathrm{m}_1=-1\) Let, \(\mathrm{m}_2=\mathrm{m}\) We know that if \(\theta\) is the angle between two lines then \(\tan \theta=\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\) \(\tan 45^{\circ}=\frac{\mathrm{m}+1}{1-\mathrm{m}}\) \(1-\mathrm{m}=\mathrm{m}+1\) \(2 \mathrm{~m}=0\) \(\mathrm{~m}=0\) and \(\quad 1=\frac{\mathrm{m}+1}{\mathrm{~m}-1}\) \(\mathrm{m}=\infty\) Then equation of line passing through \((1,1)\) is when \(\left(\mathrm{y}-\mathrm{y}_1\right)=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)\) \(\mathrm{m}=0,(\mathrm{y}-1)=0\) when \(\mathrm{m}=\infty,(\mathrm{x}-1)=0\) Hence, the equation of the lines are \(\mathrm{x}-1=0, \mathrm{y}-1=0\)
WB JEE-2010
Conic Section
119668
The incentre of an equilateral triangle is \((1,1)\) and the equation of one side is \(3 x+4 y+3=0\). Then, the equation of the circumcircle of the triangle is
1 \(x^2+y^2-2 x-2 y-2=0\)
2 \(x^2+y^2-2 x-2 y-14=0\)
3 \(x^2+y^2-2 x-2 y+2=0\)
4 \(x^2+y^2-2 x-2 y+14=0\)
Explanation:
B Given, equation \(3 \mathrm{x}+4 \mathrm{y}+3=0\) and centre \((1,1)\) we know that, Radius \((\mathrm{r})\) of incircle \(=\frac{|3 \times(1)+4 \times(1)+3|}{\sqrt{(3)^2+(4)^2}}\) \(\mathrm{r}=\frac{10}{5}=2\) For equilateral triangle circumcentre and incentre coincide. \(\sin 30^{\circ}=\frac{2}{R}\) \(\mathrm{R}=4\) Therefore, equation of circle with centre \((1,1)\) and \(\operatorname{radius}(\mathrm{R})=4\) \((x-1)^2+(y-1)^2=4^2=16\) \(x^2+y^2-2 x-2 y-14=0\)
WB JEE-2012
Conic Section
119669
If four distinct points \((2 k, 3 k),(2,0),(0,3),(0,0)\) lie on a circle, then
1 \(\mathrm{k}\lt 0\)
2 \(0\lt \mathrm{k}\lt 1\)
3 \(\mathrm{k}=1\)
4 \(\mathrm{k}>1\)
Explanation:
C Given four distinct points \((2 \mathrm{k}, 3 \mathrm{k}),(2,0),(0,3),(0,0)\) Since, join of \((2,0)\) and \((0,3)\) subtends \(90^{\circ}\) at \((0,0)\) It is diameter. Equation is \((\mathrm{x}-2)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-3)=0\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-3 \mathrm{y}=0\) \((2 \mathrm{k}, 3 \mathrm{k})\) lies on it \(4 \mathrm{k}^2+9 \mathrm{k}^2-4 \mathrm{k}-9 \mathrm{k}=0\) \(13 \mathrm{k}^2=13 \mathrm{k}\) \(\mathrm{k}=1\) Since, \(\mathrm{k} \in 0\) otherwise \((2 \mathrm{k}, 3 \mathrm{k})\) will be \((0,0)\).
WB JEE-2012
Conic Section
119670
The locus of the mid-points of chords of the circle \(x^2+y^2=1\), which subtends a right angle at the origin, is
1 \(x^2+y^2=\frac{1}{4}\)
2 \(x^2+y^2=\frac{1}{2}\)
3 \(x y=0\)
4 \(x^2-y^2=0\)
Explanation:
B Let the centre of the circle \((0,0)\) be \(\mathrm{O}\). Therefore \(\mathrm{CM}=\mathrm{R} \sin \left(\frac{\pi}{4}\right) \text { or } \mathrm{CM}=\frac{\mathrm{R}}{\sqrt{2}}\) Hence, \(\mathrm{CM}^2=\frac{\mathrm{R}^2}{2}=\frac{1}{2}\) Let the midpoint of the chord be \(\mathrm{M}(\mathrm{h}, \mathrm{k})\). Then distance formula gives \(\mathrm{CM}^2=(\mathrm{h}-0)^2+(\mathrm{k}-0)^2=\mathrm{OM}^2\) \(\mathrm{CM}^2=\mathrm{h}^2+\mathrm{k}^2\) Replacing (h, k) by ( \(\mathrm{x}, \mathrm{y})\) and \(\mathrm{R}=1\), we get the equation of the midpoint of the chord as \(\mathrm{x}^2+\mathrm{y}^2=\frac{1}{2}\)
WB JEE-2016
Conic Section
119671
The focus of the conic \(x^2-6 x+4 y+1=0\) is
1 \((2,3)\)
2 \((3,2)\)
3 \((3,1)\)
4 \((1,4)\)
Explanation:
C Given, equation of conic is \(x^2-6 x+4 y+1=0\) \(x^2-6 x+9-9+4 y+1=0\) \(x^2-3 x-3 x+9-9+4 y+1=0\) \(x(x-3)-3(x-3)+4 y-8=0\) \((x-3)^2+4 y-8=0\) \((x-3)^2=-4 y+8\) \((x-3)^2=-4(y-2)\) \((x-3)^2=4(-1)(y-2)\) \(\text { Thus, the focus is }(3,1)\)
119667
The equations of the lines through \((1,1)\) and making angles of \(45^{\circ}\) with the line \(x+y=0\) are
1 \(x-1=0, x-y=0\)
2 \(x-y=0, y-1=0\)
3 \(x+y-2=0, y-1=0\)
4 \(x-1=0, y-1=0\)
Explanation:
D Given, equation of line \(\mathrm{x}+\mathrm{y}=0\) and \(\theta=45^{\circ}\) \(y=-x\) Slope of line \(\mathrm{m}_1=-1\) Let, \(\mathrm{m}_2=\mathrm{m}\) We know that if \(\theta\) is the angle between two lines then \(\tan \theta=\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\) \(\tan 45^{\circ}=\frac{\mathrm{m}+1}{1-\mathrm{m}}\) \(1-\mathrm{m}=\mathrm{m}+1\) \(2 \mathrm{~m}=0\) \(\mathrm{~m}=0\) and \(\quad 1=\frac{\mathrm{m}+1}{\mathrm{~m}-1}\) \(\mathrm{m}=\infty\) Then equation of line passing through \((1,1)\) is when \(\left(\mathrm{y}-\mathrm{y}_1\right)=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)\) \(\mathrm{m}=0,(\mathrm{y}-1)=0\) when \(\mathrm{m}=\infty,(\mathrm{x}-1)=0\) Hence, the equation of the lines are \(\mathrm{x}-1=0, \mathrm{y}-1=0\)
WB JEE-2010
Conic Section
119668
The incentre of an equilateral triangle is \((1,1)\) and the equation of one side is \(3 x+4 y+3=0\). Then, the equation of the circumcircle of the triangle is
1 \(x^2+y^2-2 x-2 y-2=0\)
2 \(x^2+y^2-2 x-2 y-14=0\)
3 \(x^2+y^2-2 x-2 y+2=0\)
4 \(x^2+y^2-2 x-2 y+14=0\)
Explanation:
B Given, equation \(3 \mathrm{x}+4 \mathrm{y}+3=0\) and centre \((1,1)\) we know that, Radius \((\mathrm{r})\) of incircle \(=\frac{|3 \times(1)+4 \times(1)+3|}{\sqrt{(3)^2+(4)^2}}\) \(\mathrm{r}=\frac{10}{5}=2\) For equilateral triangle circumcentre and incentre coincide. \(\sin 30^{\circ}=\frac{2}{R}\) \(\mathrm{R}=4\) Therefore, equation of circle with centre \((1,1)\) and \(\operatorname{radius}(\mathrm{R})=4\) \((x-1)^2+(y-1)^2=4^2=16\) \(x^2+y^2-2 x-2 y-14=0\)
WB JEE-2012
Conic Section
119669
If four distinct points \((2 k, 3 k),(2,0),(0,3),(0,0)\) lie on a circle, then
1 \(\mathrm{k}\lt 0\)
2 \(0\lt \mathrm{k}\lt 1\)
3 \(\mathrm{k}=1\)
4 \(\mathrm{k}>1\)
Explanation:
C Given four distinct points \((2 \mathrm{k}, 3 \mathrm{k}),(2,0),(0,3),(0,0)\) Since, join of \((2,0)\) and \((0,3)\) subtends \(90^{\circ}\) at \((0,0)\) It is diameter. Equation is \((\mathrm{x}-2)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-3)=0\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-3 \mathrm{y}=0\) \((2 \mathrm{k}, 3 \mathrm{k})\) lies on it \(4 \mathrm{k}^2+9 \mathrm{k}^2-4 \mathrm{k}-9 \mathrm{k}=0\) \(13 \mathrm{k}^2=13 \mathrm{k}\) \(\mathrm{k}=1\) Since, \(\mathrm{k} \in 0\) otherwise \((2 \mathrm{k}, 3 \mathrm{k})\) will be \((0,0)\).
WB JEE-2012
Conic Section
119670
The locus of the mid-points of chords of the circle \(x^2+y^2=1\), which subtends a right angle at the origin, is
1 \(x^2+y^2=\frac{1}{4}\)
2 \(x^2+y^2=\frac{1}{2}\)
3 \(x y=0\)
4 \(x^2-y^2=0\)
Explanation:
B Let the centre of the circle \((0,0)\) be \(\mathrm{O}\). Therefore \(\mathrm{CM}=\mathrm{R} \sin \left(\frac{\pi}{4}\right) \text { or } \mathrm{CM}=\frac{\mathrm{R}}{\sqrt{2}}\) Hence, \(\mathrm{CM}^2=\frac{\mathrm{R}^2}{2}=\frac{1}{2}\) Let the midpoint of the chord be \(\mathrm{M}(\mathrm{h}, \mathrm{k})\). Then distance formula gives \(\mathrm{CM}^2=(\mathrm{h}-0)^2+(\mathrm{k}-0)^2=\mathrm{OM}^2\) \(\mathrm{CM}^2=\mathrm{h}^2+\mathrm{k}^2\) Replacing (h, k) by ( \(\mathrm{x}, \mathrm{y})\) and \(\mathrm{R}=1\), we get the equation of the midpoint of the chord as \(\mathrm{x}^2+\mathrm{y}^2=\frac{1}{2}\)
WB JEE-2016
Conic Section
119671
The focus of the conic \(x^2-6 x+4 y+1=0\) is
1 \((2,3)\)
2 \((3,2)\)
3 \((3,1)\)
4 \((1,4)\)
Explanation:
C Given, equation of conic is \(x^2-6 x+4 y+1=0\) \(x^2-6 x+9-9+4 y+1=0\) \(x^2-3 x-3 x+9-9+4 y+1=0\) \(x(x-3)-3(x-3)+4 y-8=0\) \((x-3)^2+4 y-8=0\) \((x-3)^2=-4 y+8\) \((x-3)^2=-4(y-2)\) \((x-3)^2=4(-1)(y-2)\) \(\text { Thus, the focus is }(3,1)\)
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Conic Section
119667
The equations of the lines through \((1,1)\) and making angles of \(45^{\circ}\) with the line \(x+y=0\) are
1 \(x-1=0, x-y=0\)
2 \(x-y=0, y-1=0\)
3 \(x+y-2=0, y-1=0\)
4 \(x-1=0, y-1=0\)
Explanation:
D Given, equation of line \(\mathrm{x}+\mathrm{y}=0\) and \(\theta=45^{\circ}\) \(y=-x\) Slope of line \(\mathrm{m}_1=-1\) Let, \(\mathrm{m}_2=\mathrm{m}\) We know that if \(\theta\) is the angle between two lines then \(\tan \theta=\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\) \(\tan 45^{\circ}=\frac{\mathrm{m}+1}{1-\mathrm{m}}\) \(1-\mathrm{m}=\mathrm{m}+1\) \(2 \mathrm{~m}=0\) \(\mathrm{~m}=0\) and \(\quad 1=\frac{\mathrm{m}+1}{\mathrm{~m}-1}\) \(\mathrm{m}=\infty\) Then equation of line passing through \((1,1)\) is when \(\left(\mathrm{y}-\mathrm{y}_1\right)=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)\) \(\mathrm{m}=0,(\mathrm{y}-1)=0\) when \(\mathrm{m}=\infty,(\mathrm{x}-1)=0\) Hence, the equation of the lines are \(\mathrm{x}-1=0, \mathrm{y}-1=0\)
WB JEE-2010
Conic Section
119668
The incentre of an equilateral triangle is \((1,1)\) and the equation of one side is \(3 x+4 y+3=0\). Then, the equation of the circumcircle of the triangle is
1 \(x^2+y^2-2 x-2 y-2=0\)
2 \(x^2+y^2-2 x-2 y-14=0\)
3 \(x^2+y^2-2 x-2 y+2=0\)
4 \(x^2+y^2-2 x-2 y+14=0\)
Explanation:
B Given, equation \(3 \mathrm{x}+4 \mathrm{y}+3=0\) and centre \((1,1)\) we know that, Radius \((\mathrm{r})\) of incircle \(=\frac{|3 \times(1)+4 \times(1)+3|}{\sqrt{(3)^2+(4)^2}}\) \(\mathrm{r}=\frac{10}{5}=2\) For equilateral triangle circumcentre and incentre coincide. \(\sin 30^{\circ}=\frac{2}{R}\) \(\mathrm{R}=4\) Therefore, equation of circle with centre \((1,1)\) and \(\operatorname{radius}(\mathrm{R})=4\) \((x-1)^2+(y-1)^2=4^2=16\) \(x^2+y^2-2 x-2 y-14=0\)
WB JEE-2012
Conic Section
119669
If four distinct points \((2 k, 3 k),(2,0),(0,3),(0,0)\) lie on a circle, then
1 \(\mathrm{k}\lt 0\)
2 \(0\lt \mathrm{k}\lt 1\)
3 \(\mathrm{k}=1\)
4 \(\mathrm{k}>1\)
Explanation:
C Given four distinct points \((2 \mathrm{k}, 3 \mathrm{k}),(2,0),(0,3),(0,0)\) Since, join of \((2,0)\) and \((0,3)\) subtends \(90^{\circ}\) at \((0,0)\) It is diameter. Equation is \((\mathrm{x}-2)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-3)=0\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-3 \mathrm{y}=0\) \((2 \mathrm{k}, 3 \mathrm{k})\) lies on it \(4 \mathrm{k}^2+9 \mathrm{k}^2-4 \mathrm{k}-9 \mathrm{k}=0\) \(13 \mathrm{k}^2=13 \mathrm{k}\) \(\mathrm{k}=1\) Since, \(\mathrm{k} \in 0\) otherwise \((2 \mathrm{k}, 3 \mathrm{k})\) will be \((0,0)\).
WB JEE-2012
Conic Section
119670
The locus of the mid-points of chords of the circle \(x^2+y^2=1\), which subtends a right angle at the origin, is
1 \(x^2+y^2=\frac{1}{4}\)
2 \(x^2+y^2=\frac{1}{2}\)
3 \(x y=0\)
4 \(x^2-y^2=0\)
Explanation:
B Let the centre of the circle \((0,0)\) be \(\mathrm{O}\). Therefore \(\mathrm{CM}=\mathrm{R} \sin \left(\frac{\pi}{4}\right) \text { or } \mathrm{CM}=\frac{\mathrm{R}}{\sqrt{2}}\) Hence, \(\mathrm{CM}^2=\frac{\mathrm{R}^2}{2}=\frac{1}{2}\) Let the midpoint of the chord be \(\mathrm{M}(\mathrm{h}, \mathrm{k})\). Then distance formula gives \(\mathrm{CM}^2=(\mathrm{h}-0)^2+(\mathrm{k}-0)^2=\mathrm{OM}^2\) \(\mathrm{CM}^2=\mathrm{h}^2+\mathrm{k}^2\) Replacing (h, k) by ( \(\mathrm{x}, \mathrm{y})\) and \(\mathrm{R}=1\), we get the equation of the midpoint of the chord as \(\mathrm{x}^2+\mathrm{y}^2=\frac{1}{2}\)
WB JEE-2016
Conic Section
119671
The focus of the conic \(x^2-6 x+4 y+1=0\) is
1 \((2,3)\)
2 \((3,2)\)
3 \((3,1)\)
4 \((1,4)\)
Explanation:
C Given, equation of conic is \(x^2-6 x+4 y+1=0\) \(x^2-6 x+9-9+4 y+1=0\) \(x^2-3 x-3 x+9-9+4 y+1=0\) \(x(x-3)-3(x-3)+4 y-8=0\) \((x-3)^2+4 y-8=0\) \((x-3)^2=-4 y+8\) \((x-3)^2=-4(y-2)\) \((x-3)^2=4(-1)(y-2)\) \(\text { Thus, the focus is }(3,1)\)
119667
The equations of the lines through \((1,1)\) and making angles of \(45^{\circ}\) with the line \(x+y=0\) are
1 \(x-1=0, x-y=0\)
2 \(x-y=0, y-1=0\)
3 \(x+y-2=0, y-1=0\)
4 \(x-1=0, y-1=0\)
Explanation:
D Given, equation of line \(\mathrm{x}+\mathrm{y}=0\) and \(\theta=45^{\circ}\) \(y=-x\) Slope of line \(\mathrm{m}_1=-1\) Let, \(\mathrm{m}_2=\mathrm{m}\) We know that if \(\theta\) is the angle between two lines then \(\tan \theta=\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\) \(\tan 45^{\circ}=\frac{\mathrm{m}+1}{1-\mathrm{m}}\) \(1-\mathrm{m}=\mathrm{m}+1\) \(2 \mathrm{~m}=0\) \(\mathrm{~m}=0\) and \(\quad 1=\frac{\mathrm{m}+1}{\mathrm{~m}-1}\) \(\mathrm{m}=\infty\) Then equation of line passing through \((1,1)\) is when \(\left(\mathrm{y}-\mathrm{y}_1\right)=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)\) \(\mathrm{m}=0,(\mathrm{y}-1)=0\) when \(\mathrm{m}=\infty,(\mathrm{x}-1)=0\) Hence, the equation of the lines are \(\mathrm{x}-1=0, \mathrm{y}-1=0\)
WB JEE-2010
Conic Section
119668
The incentre of an equilateral triangle is \((1,1)\) and the equation of one side is \(3 x+4 y+3=0\). Then, the equation of the circumcircle of the triangle is
1 \(x^2+y^2-2 x-2 y-2=0\)
2 \(x^2+y^2-2 x-2 y-14=0\)
3 \(x^2+y^2-2 x-2 y+2=0\)
4 \(x^2+y^2-2 x-2 y+14=0\)
Explanation:
B Given, equation \(3 \mathrm{x}+4 \mathrm{y}+3=0\) and centre \((1,1)\) we know that, Radius \((\mathrm{r})\) of incircle \(=\frac{|3 \times(1)+4 \times(1)+3|}{\sqrt{(3)^2+(4)^2}}\) \(\mathrm{r}=\frac{10}{5}=2\) For equilateral triangle circumcentre and incentre coincide. \(\sin 30^{\circ}=\frac{2}{R}\) \(\mathrm{R}=4\) Therefore, equation of circle with centre \((1,1)\) and \(\operatorname{radius}(\mathrm{R})=4\) \((x-1)^2+(y-1)^2=4^2=16\) \(x^2+y^2-2 x-2 y-14=0\)
WB JEE-2012
Conic Section
119669
If four distinct points \((2 k, 3 k),(2,0),(0,3),(0,0)\) lie on a circle, then
1 \(\mathrm{k}\lt 0\)
2 \(0\lt \mathrm{k}\lt 1\)
3 \(\mathrm{k}=1\)
4 \(\mathrm{k}>1\)
Explanation:
C Given four distinct points \((2 \mathrm{k}, 3 \mathrm{k}),(2,0),(0,3),(0,0)\) Since, join of \((2,0)\) and \((0,3)\) subtends \(90^{\circ}\) at \((0,0)\) It is diameter. Equation is \((\mathrm{x}-2)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-3)=0\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-3 \mathrm{y}=0\) \((2 \mathrm{k}, 3 \mathrm{k})\) lies on it \(4 \mathrm{k}^2+9 \mathrm{k}^2-4 \mathrm{k}-9 \mathrm{k}=0\) \(13 \mathrm{k}^2=13 \mathrm{k}\) \(\mathrm{k}=1\) Since, \(\mathrm{k} \in 0\) otherwise \((2 \mathrm{k}, 3 \mathrm{k})\) will be \((0,0)\).
WB JEE-2012
Conic Section
119670
The locus of the mid-points of chords of the circle \(x^2+y^2=1\), which subtends a right angle at the origin, is
1 \(x^2+y^2=\frac{1}{4}\)
2 \(x^2+y^2=\frac{1}{2}\)
3 \(x y=0\)
4 \(x^2-y^2=0\)
Explanation:
B Let the centre of the circle \((0,0)\) be \(\mathrm{O}\). Therefore \(\mathrm{CM}=\mathrm{R} \sin \left(\frac{\pi}{4}\right) \text { or } \mathrm{CM}=\frac{\mathrm{R}}{\sqrt{2}}\) Hence, \(\mathrm{CM}^2=\frac{\mathrm{R}^2}{2}=\frac{1}{2}\) Let the midpoint of the chord be \(\mathrm{M}(\mathrm{h}, \mathrm{k})\). Then distance formula gives \(\mathrm{CM}^2=(\mathrm{h}-0)^2+(\mathrm{k}-0)^2=\mathrm{OM}^2\) \(\mathrm{CM}^2=\mathrm{h}^2+\mathrm{k}^2\) Replacing (h, k) by ( \(\mathrm{x}, \mathrm{y})\) and \(\mathrm{R}=1\), we get the equation of the midpoint of the chord as \(\mathrm{x}^2+\mathrm{y}^2=\frac{1}{2}\)
WB JEE-2016
Conic Section
119671
The focus of the conic \(x^2-6 x+4 y+1=0\) is
1 \((2,3)\)
2 \((3,2)\)
3 \((3,1)\)
4 \((1,4)\)
Explanation:
C Given, equation of conic is \(x^2-6 x+4 y+1=0\) \(x^2-6 x+9-9+4 y+1=0\) \(x^2-3 x-3 x+9-9+4 y+1=0\) \(x(x-3)-3(x-3)+4 y-8=0\) \((x-3)^2+4 y-8=0\) \((x-3)^2=-4 y+8\) \((x-3)^2=-4(y-2)\) \((x-3)^2=4(-1)(y-2)\) \(\text { Thus, the focus is }(3,1)\)
119667
The equations of the lines through \((1,1)\) and making angles of \(45^{\circ}\) with the line \(x+y=0\) are
1 \(x-1=0, x-y=0\)
2 \(x-y=0, y-1=0\)
3 \(x+y-2=0, y-1=0\)
4 \(x-1=0, y-1=0\)
Explanation:
D Given, equation of line \(\mathrm{x}+\mathrm{y}=0\) and \(\theta=45^{\circ}\) \(y=-x\) Slope of line \(\mathrm{m}_1=-1\) Let, \(\mathrm{m}_2=\mathrm{m}\) We know that if \(\theta\) is the angle between two lines then \(\tan \theta=\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\) \(\tan 45^{\circ}=\frac{\mathrm{m}+1}{1-\mathrm{m}}\) \(1-\mathrm{m}=\mathrm{m}+1\) \(2 \mathrm{~m}=0\) \(\mathrm{~m}=0\) and \(\quad 1=\frac{\mathrm{m}+1}{\mathrm{~m}-1}\) \(\mathrm{m}=\infty\) Then equation of line passing through \((1,1)\) is when \(\left(\mathrm{y}-\mathrm{y}_1\right)=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)\) \(\mathrm{m}=0,(\mathrm{y}-1)=0\) when \(\mathrm{m}=\infty,(\mathrm{x}-1)=0\) Hence, the equation of the lines are \(\mathrm{x}-1=0, \mathrm{y}-1=0\)
WB JEE-2010
Conic Section
119668
The incentre of an equilateral triangle is \((1,1)\) and the equation of one side is \(3 x+4 y+3=0\). Then, the equation of the circumcircle of the triangle is
1 \(x^2+y^2-2 x-2 y-2=0\)
2 \(x^2+y^2-2 x-2 y-14=0\)
3 \(x^2+y^2-2 x-2 y+2=0\)
4 \(x^2+y^2-2 x-2 y+14=0\)
Explanation:
B Given, equation \(3 \mathrm{x}+4 \mathrm{y}+3=0\) and centre \((1,1)\) we know that, Radius \((\mathrm{r})\) of incircle \(=\frac{|3 \times(1)+4 \times(1)+3|}{\sqrt{(3)^2+(4)^2}}\) \(\mathrm{r}=\frac{10}{5}=2\) For equilateral triangle circumcentre and incentre coincide. \(\sin 30^{\circ}=\frac{2}{R}\) \(\mathrm{R}=4\) Therefore, equation of circle with centre \((1,1)\) and \(\operatorname{radius}(\mathrm{R})=4\) \((x-1)^2+(y-1)^2=4^2=16\) \(x^2+y^2-2 x-2 y-14=0\)
WB JEE-2012
Conic Section
119669
If four distinct points \((2 k, 3 k),(2,0),(0,3),(0,0)\) lie on a circle, then
1 \(\mathrm{k}\lt 0\)
2 \(0\lt \mathrm{k}\lt 1\)
3 \(\mathrm{k}=1\)
4 \(\mathrm{k}>1\)
Explanation:
C Given four distinct points \((2 \mathrm{k}, 3 \mathrm{k}),(2,0),(0,3),(0,0)\) Since, join of \((2,0)\) and \((0,3)\) subtends \(90^{\circ}\) at \((0,0)\) It is diameter. Equation is \((\mathrm{x}-2)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-3)=0\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-3 \mathrm{y}=0\) \((2 \mathrm{k}, 3 \mathrm{k})\) lies on it \(4 \mathrm{k}^2+9 \mathrm{k}^2-4 \mathrm{k}-9 \mathrm{k}=0\) \(13 \mathrm{k}^2=13 \mathrm{k}\) \(\mathrm{k}=1\) Since, \(\mathrm{k} \in 0\) otherwise \((2 \mathrm{k}, 3 \mathrm{k})\) will be \((0,0)\).
WB JEE-2012
Conic Section
119670
The locus of the mid-points of chords of the circle \(x^2+y^2=1\), which subtends a right angle at the origin, is
1 \(x^2+y^2=\frac{1}{4}\)
2 \(x^2+y^2=\frac{1}{2}\)
3 \(x y=0\)
4 \(x^2-y^2=0\)
Explanation:
B Let the centre of the circle \((0,0)\) be \(\mathrm{O}\). Therefore \(\mathrm{CM}=\mathrm{R} \sin \left(\frac{\pi}{4}\right) \text { or } \mathrm{CM}=\frac{\mathrm{R}}{\sqrt{2}}\) Hence, \(\mathrm{CM}^2=\frac{\mathrm{R}^2}{2}=\frac{1}{2}\) Let the midpoint of the chord be \(\mathrm{M}(\mathrm{h}, \mathrm{k})\). Then distance formula gives \(\mathrm{CM}^2=(\mathrm{h}-0)^2+(\mathrm{k}-0)^2=\mathrm{OM}^2\) \(\mathrm{CM}^2=\mathrm{h}^2+\mathrm{k}^2\) Replacing (h, k) by ( \(\mathrm{x}, \mathrm{y})\) and \(\mathrm{R}=1\), we get the equation of the midpoint of the chord as \(\mathrm{x}^2+\mathrm{y}^2=\frac{1}{2}\)
WB JEE-2016
Conic Section
119671
The focus of the conic \(x^2-6 x+4 y+1=0\) is
1 \((2,3)\)
2 \((3,2)\)
3 \((3,1)\)
4 \((1,4)\)
Explanation:
C Given, equation of conic is \(x^2-6 x+4 y+1=0\) \(x^2-6 x+9-9+4 y+1=0\) \(x^2-3 x-3 x+9-9+4 y+1=0\) \(x(x-3)-3(x-3)+4 y-8=0\) \((x-3)^2+4 y-8=0\) \((x-3)^2=-4 y+8\) \((x-3)^2=-4(y-2)\) \((x-3)^2=4(-1)(y-2)\) \(\text { Thus, the focus is }(3,1)\)
