119631
The equation of the circle which passes through the point \((4,5)\) and has its centre at \((2\), 2) is
1 \((x-2)+(y-2)=13\)
2 \((x-2)^2+(y-2)^2=13\)
3 \((x)^2+(y)^2=13\)
4 \((x-4)^2+(y-5)^2=13\)
Explanation:
B As the circle is passing through the point \((4,5)\) and its centre is \((2,2)\) So its radius is \(r=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) \(r=\sqrt{(4-2)^2+(5-2)^2}\) \(r=\sqrt{4+9}=\sqrt{13}\) \(r=\sqrt{13}\)\(\therefore\) The required equation is \(-(x-2)^2+(y-2)^2=13\)
BITSAT-2018
Conic Section
119691
Let \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) be three points on a circle of radius 1 such that \(\angle \mathrm{ACB}=\frac{\pi}{4}\). Then the length of the side \(A B\) is
1 \(\sqrt{3}\)
2 \(\frac{4}{3}\)
3 \(\frac{3}{\sqrt{2}}\)
4 \(\sqrt{2}\)
Explanation:
D Given that, \(\mathrm{r}=1, \angle \mathrm{ACB}=\frac{\pi}{4}\) Let \(O\) be the centre of the circle In \(\triangle \mathrm{OAB}\) \(\mathrm{AB}=\sqrt{2} \mathrm{r}\) And \(\mathrm{r}=1\) \(\mathrm{AB}=\sqrt{2} \times 1\) \(\mathrm{AB}=\sqrt{2}\)
KVPY SA-2020
Conic Section
119700
The centre of the circle \(2 x^2+2 y^2+\frac{3}{2} x+9=0\) is
1 \(\left(\frac{3}{8}, 0\right)\)
2 \(\left(-\frac{3}{8}, 0\right)\)
3 \(\left(0, \frac{3}{8}\right)\)
4 \(\left(0,-\frac{3}{8}\right)\)
Explanation:
B Given, \(2 x^2+2 y^2+\frac{3}{2} x+9=0\) \(\mathrm{x}^2+\mathrm{y}^2+\frac{3}{4} \mathrm{x}+\frac{9}{2}=0\) \(2 \mathrm{~g}=\frac{3}{4}, 2 \mathrm{f}=0\) \(\mathrm{~g}=\frac{3}{8} \mathrm{f}=0\)So, centre is \((-\mathrm{g},-\mathrm{f})=\left(-\frac{3}{8}, 0\right)\)
GUJCET-2011
Conic Section
119722
The centre and radius of the circle \(x^2+y^2-4 x\) \(+2 \mathbf{y}=0\) are
1 \((2,-1)\) and 5
2 \((4,2)\) and \(\sqrt{20}\)
3 \((2,-1)\) and \(\sqrt{5}\)
4 \((-2,1)\) and 5
5 \((-2,1)\) and \(\sqrt{5}\)
Explanation:
C Given, \(x^2+y^2-4 x+2 y=0\) Here, \(2 g=-4,2 f=2, c=0\) \(g=-2, f=1\) Then centre, \((-\mathrm{g},-\mathrm{f})=(2,-1)\) And, \(\mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(\mathrm{r}=\sqrt{(-2)^2+(1)^2-0}\) \(\mathrm{r}=\sqrt{4+1}=\sqrt{5}\)
Kerala CEE-2020
Conic Section
119647
Consider the circles \(x^2+(y-1)^2=9\), \((x-1)^2+y^2=25\). They are such that
1 these circles touch each other
2 One of these circles lies entirely inside the other
3 each of these circles lies outside the other
4 they intersect in two points
Explanation:
B The circle are \(x^2+(y-1)^2=9\) \(\text { And, }\) \((x-1)^2+(y)^2=25\) \(c_1=(0,1), c_2=(1,0)\) \(\quad r_1=3, \quad r_2=5\) \(\mathrm{c}_1 \mathrm{c}_2=\sqrt{1^2+1^2}=\sqrt{2}\) \(\mathrm{r}_2-\mathrm{r}_1=2\) \(\mathrm{c}_1 \mathrm{c}_2\lt \left(\mathrm{r}_2-\mathrm{r}_1\right)\) \(\sqrt{2}\lt \mathrm{r}_2-\mathrm{r}_1\)Therefore, one circle lies entirely inside the other circle.
119631
The equation of the circle which passes through the point \((4,5)\) and has its centre at \((2\), 2) is
1 \((x-2)+(y-2)=13\)
2 \((x-2)^2+(y-2)^2=13\)
3 \((x)^2+(y)^2=13\)
4 \((x-4)^2+(y-5)^2=13\)
Explanation:
B As the circle is passing through the point \((4,5)\) and its centre is \((2,2)\) So its radius is \(r=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) \(r=\sqrt{(4-2)^2+(5-2)^2}\) \(r=\sqrt{4+9}=\sqrt{13}\) \(r=\sqrt{13}\)\(\therefore\) The required equation is \(-(x-2)^2+(y-2)^2=13\)
BITSAT-2018
Conic Section
119691
Let \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) be three points on a circle of radius 1 such that \(\angle \mathrm{ACB}=\frac{\pi}{4}\). Then the length of the side \(A B\) is
1 \(\sqrt{3}\)
2 \(\frac{4}{3}\)
3 \(\frac{3}{\sqrt{2}}\)
4 \(\sqrt{2}\)
Explanation:
D Given that, \(\mathrm{r}=1, \angle \mathrm{ACB}=\frac{\pi}{4}\) Let \(O\) be the centre of the circle In \(\triangle \mathrm{OAB}\) \(\mathrm{AB}=\sqrt{2} \mathrm{r}\) And \(\mathrm{r}=1\) \(\mathrm{AB}=\sqrt{2} \times 1\) \(\mathrm{AB}=\sqrt{2}\)
KVPY SA-2020
Conic Section
119700
The centre of the circle \(2 x^2+2 y^2+\frac{3}{2} x+9=0\) is
1 \(\left(\frac{3}{8}, 0\right)\)
2 \(\left(-\frac{3}{8}, 0\right)\)
3 \(\left(0, \frac{3}{8}\right)\)
4 \(\left(0,-\frac{3}{8}\right)\)
Explanation:
B Given, \(2 x^2+2 y^2+\frac{3}{2} x+9=0\) \(\mathrm{x}^2+\mathrm{y}^2+\frac{3}{4} \mathrm{x}+\frac{9}{2}=0\) \(2 \mathrm{~g}=\frac{3}{4}, 2 \mathrm{f}=0\) \(\mathrm{~g}=\frac{3}{8} \mathrm{f}=0\)So, centre is \((-\mathrm{g},-\mathrm{f})=\left(-\frac{3}{8}, 0\right)\)
GUJCET-2011
Conic Section
119722
The centre and radius of the circle \(x^2+y^2-4 x\) \(+2 \mathbf{y}=0\) are
1 \((2,-1)\) and 5
2 \((4,2)\) and \(\sqrt{20}\)
3 \((2,-1)\) and \(\sqrt{5}\)
4 \((-2,1)\) and 5
5 \((-2,1)\) and \(\sqrt{5}\)
Explanation:
C Given, \(x^2+y^2-4 x+2 y=0\) Here, \(2 g=-4,2 f=2, c=0\) \(g=-2, f=1\) Then centre, \((-\mathrm{g},-\mathrm{f})=(2,-1)\) And, \(\mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(\mathrm{r}=\sqrt{(-2)^2+(1)^2-0}\) \(\mathrm{r}=\sqrt{4+1}=\sqrt{5}\)
Kerala CEE-2020
Conic Section
119647
Consider the circles \(x^2+(y-1)^2=9\), \((x-1)^2+y^2=25\). They are such that
1 these circles touch each other
2 One of these circles lies entirely inside the other
3 each of these circles lies outside the other
4 they intersect in two points
Explanation:
B The circle are \(x^2+(y-1)^2=9\) \(\text { And, }\) \((x-1)^2+(y)^2=25\) \(c_1=(0,1), c_2=(1,0)\) \(\quad r_1=3, \quad r_2=5\) \(\mathrm{c}_1 \mathrm{c}_2=\sqrt{1^2+1^2}=\sqrt{2}\) \(\mathrm{r}_2-\mathrm{r}_1=2\) \(\mathrm{c}_1 \mathrm{c}_2\lt \left(\mathrm{r}_2-\mathrm{r}_1\right)\) \(\sqrt{2}\lt \mathrm{r}_2-\mathrm{r}_1\)Therefore, one circle lies entirely inside the other circle.
119631
The equation of the circle which passes through the point \((4,5)\) and has its centre at \((2\), 2) is
1 \((x-2)+(y-2)=13\)
2 \((x-2)^2+(y-2)^2=13\)
3 \((x)^2+(y)^2=13\)
4 \((x-4)^2+(y-5)^2=13\)
Explanation:
B As the circle is passing through the point \((4,5)\) and its centre is \((2,2)\) So its radius is \(r=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) \(r=\sqrt{(4-2)^2+(5-2)^2}\) \(r=\sqrt{4+9}=\sqrt{13}\) \(r=\sqrt{13}\)\(\therefore\) The required equation is \(-(x-2)^2+(y-2)^2=13\)
BITSAT-2018
Conic Section
119691
Let \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) be three points on a circle of radius 1 such that \(\angle \mathrm{ACB}=\frac{\pi}{4}\). Then the length of the side \(A B\) is
1 \(\sqrt{3}\)
2 \(\frac{4}{3}\)
3 \(\frac{3}{\sqrt{2}}\)
4 \(\sqrt{2}\)
Explanation:
D Given that, \(\mathrm{r}=1, \angle \mathrm{ACB}=\frac{\pi}{4}\) Let \(O\) be the centre of the circle In \(\triangle \mathrm{OAB}\) \(\mathrm{AB}=\sqrt{2} \mathrm{r}\) And \(\mathrm{r}=1\) \(\mathrm{AB}=\sqrt{2} \times 1\) \(\mathrm{AB}=\sqrt{2}\)
KVPY SA-2020
Conic Section
119700
The centre of the circle \(2 x^2+2 y^2+\frac{3}{2} x+9=0\) is
1 \(\left(\frac{3}{8}, 0\right)\)
2 \(\left(-\frac{3}{8}, 0\right)\)
3 \(\left(0, \frac{3}{8}\right)\)
4 \(\left(0,-\frac{3}{8}\right)\)
Explanation:
B Given, \(2 x^2+2 y^2+\frac{3}{2} x+9=0\) \(\mathrm{x}^2+\mathrm{y}^2+\frac{3}{4} \mathrm{x}+\frac{9}{2}=0\) \(2 \mathrm{~g}=\frac{3}{4}, 2 \mathrm{f}=0\) \(\mathrm{~g}=\frac{3}{8} \mathrm{f}=0\)So, centre is \((-\mathrm{g},-\mathrm{f})=\left(-\frac{3}{8}, 0\right)\)
GUJCET-2011
Conic Section
119722
The centre and radius of the circle \(x^2+y^2-4 x\) \(+2 \mathbf{y}=0\) are
1 \((2,-1)\) and 5
2 \((4,2)\) and \(\sqrt{20}\)
3 \((2,-1)\) and \(\sqrt{5}\)
4 \((-2,1)\) and 5
5 \((-2,1)\) and \(\sqrt{5}\)
Explanation:
C Given, \(x^2+y^2-4 x+2 y=0\) Here, \(2 g=-4,2 f=2, c=0\) \(g=-2, f=1\) Then centre, \((-\mathrm{g},-\mathrm{f})=(2,-1)\) And, \(\mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(\mathrm{r}=\sqrt{(-2)^2+(1)^2-0}\) \(\mathrm{r}=\sqrt{4+1}=\sqrt{5}\)
Kerala CEE-2020
Conic Section
119647
Consider the circles \(x^2+(y-1)^2=9\), \((x-1)^2+y^2=25\). They are such that
1 these circles touch each other
2 One of these circles lies entirely inside the other
3 each of these circles lies outside the other
4 they intersect in two points
Explanation:
B The circle are \(x^2+(y-1)^2=9\) \(\text { And, }\) \((x-1)^2+(y)^2=25\) \(c_1=(0,1), c_2=(1,0)\) \(\quad r_1=3, \quad r_2=5\) \(\mathrm{c}_1 \mathrm{c}_2=\sqrt{1^2+1^2}=\sqrt{2}\) \(\mathrm{r}_2-\mathrm{r}_1=2\) \(\mathrm{c}_1 \mathrm{c}_2\lt \left(\mathrm{r}_2-\mathrm{r}_1\right)\) \(\sqrt{2}\lt \mathrm{r}_2-\mathrm{r}_1\)Therefore, one circle lies entirely inside the other circle.
119631
The equation of the circle which passes through the point \((4,5)\) and has its centre at \((2\), 2) is
1 \((x-2)+(y-2)=13\)
2 \((x-2)^2+(y-2)^2=13\)
3 \((x)^2+(y)^2=13\)
4 \((x-4)^2+(y-5)^2=13\)
Explanation:
B As the circle is passing through the point \((4,5)\) and its centre is \((2,2)\) So its radius is \(r=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) \(r=\sqrt{(4-2)^2+(5-2)^2}\) \(r=\sqrt{4+9}=\sqrt{13}\) \(r=\sqrt{13}\)\(\therefore\) The required equation is \(-(x-2)^2+(y-2)^2=13\)
BITSAT-2018
Conic Section
119691
Let \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) be three points on a circle of radius 1 such that \(\angle \mathrm{ACB}=\frac{\pi}{4}\). Then the length of the side \(A B\) is
1 \(\sqrt{3}\)
2 \(\frac{4}{3}\)
3 \(\frac{3}{\sqrt{2}}\)
4 \(\sqrt{2}\)
Explanation:
D Given that, \(\mathrm{r}=1, \angle \mathrm{ACB}=\frac{\pi}{4}\) Let \(O\) be the centre of the circle In \(\triangle \mathrm{OAB}\) \(\mathrm{AB}=\sqrt{2} \mathrm{r}\) And \(\mathrm{r}=1\) \(\mathrm{AB}=\sqrt{2} \times 1\) \(\mathrm{AB}=\sqrt{2}\)
KVPY SA-2020
Conic Section
119700
The centre of the circle \(2 x^2+2 y^2+\frac{3}{2} x+9=0\) is
1 \(\left(\frac{3}{8}, 0\right)\)
2 \(\left(-\frac{3}{8}, 0\right)\)
3 \(\left(0, \frac{3}{8}\right)\)
4 \(\left(0,-\frac{3}{8}\right)\)
Explanation:
B Given, \(2 x^2+2 y^2+\frac{3}{2} x+9=0\) \(\mathrm{x}^2+\mathrm{y}^2+\frac{3}{4} \mathrm{x}+\frac{9}{2}=0\) \(2 \mathrm{~g}=\frac{3}{4}, 2 \mathrm{f}=0\) \(\mathrm{~g}=\frac{3}{8} \mathrm{f}=0\)So, centre is \((-\mathrm{g},-\mathrm{f})=\left(-\frac{3}{8}, 0\right)\)
GUJCET-2011
Conic Section
119722
The centre and radius of the circle \(x^2+y^2-4 x\) \(+2 \mathbf{y}=0\) are
1 \((2,-1)\) and 5
2 \((4,2)\) and \(\sqrt{20}\)
3 \((2,-1)\) and \(\sqrt{5}\)
4 \((-2,1)\) and 5
5 \((-2,1)\) and \(\sqrt{5}\)
Explanation:
C Given, \(x^2+y^2-4 x+2 y=0\) Here, \(2 g=-4,2 f=2, c=0\) \(g=-2, f=1\) Then centre, \((-\mathrm{g},-\mathrm{f})=(2,-1)\) And, \(\mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(\mathrm{r}=\sqrt{(-2)^2+(1)^2-0}\) \(\mathrm{r}=\sqrt{4+1}=\sqrt{5}\)
Kerala CEE-2020
Conic Section
119647
Consider the circles \(x^2+(y-1)^2=9\), \((x-1)^2+y^2=25\). They are such that
1 these circles touch each other
2 One of these circles lies entirely inside the other
3 each of these circles lies outside the other
4 they intersect in two points
Explanation:
B The circle are \(x^2+(y-1)^2=9\) \(\text { And, }\) \((x-1)^2+(y)^2=25\) \(c_1=(0,1), c_2=(1,0)\) \(\quad r_1=3, \quad r_2=5\) \(\mathrm{c}_1 \mathrm{c}_2=\sqrt{1^2+1^2}=\sqrt{2}\) \(\mathrm{r}_2-\mathrm{r}_1=2\) \(\mathrm{c}_1 \mathrm{c}_2\lt \left(\mathrm{r}_2-\mathrm{r}_1\right)\) \(\sqrt{2}\lt \mathrm{r}_2-\mathrm{r}_1\)Therefore, one circle lies entirely inside the other circle.
119631
The equation of the circle which passes through the point \((4,5)\) and has its centre at \((2\), 2) is
1 \((x-2)+(y-2)=13\)
2 \((x-2)^2+(y-2)^2=13\)
3 \((x)^2+(y)^2=13\)
4 \((x-4)^2+(y-5)^2=13\)
Explanation:
B As the circle is passing through the point \((4,5)\) and its centre is \((2,2)\) So its radius is \(r=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) \(r=\sqrt{(4-2)^2+(5-2)^2}\) \(r=\sqrt{4+9}=\sqrt{13}\) \(r=\sqrt{13}\)\(\therefore\) The required equation is \(-(x-2)^2+(y-2)^2=13\)
BITSAT-2018
Conic Section
119691
Let \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) be three points on a circle of radius 1 such that \(\angle \mathrm{ACB}=\frac{\pi}{4}\). Then the length of the side \(A B\) is
1 \(\sqrt{3}\)
2 \(\frac{4}{3}\)
3 \(\frac{3}{\sqrt{2}}\)
4 \(\sqrt{2}\)
Explanation:
D Given that, \(\mathrm{r}=1, \angle \mathrm{ACB}=\frac{\pi}{4}\) Let \(O\) be the centre of the circle In \(\triangle \mathrm{OAB}\) \(\mathrm{AB}=\sqrt{2} \mathrm{r}\) And \(\mathrm{r}=1\) \(\mathrm{AB}=\sqrt{2} \times 1\) \(\mathrm{AB}=\sqrt{2}\)
KVPY SA-2020
Conic Section
119700
The centre of the circle \(2 x^2+2 y^2+\frac{3}{2} x+9=0\) is
1 \(\left(\frac{3}{8}, 0\right)\)
2 \(\left(-\frac{3}{8}, 0\right)\)
3 \(\left(0, \frac{3}{8}\right)\)
4 \(\left(0,-\frac{3}{8}\right)\)
Explanation:
B Given, \(2 x^2+2 y^2+\frac{3}{2} x+9=0\) \(\mathrm{x}^2+\mathrm{y}^2+\frac{3}{4} \mathrm{x}+\frac{9}{2}=0\) \(2 \mathrm{~g}=\frac{3}{4}, 2 \mathrm{f}=0\) \(\mathrm{~g}=\frac{3}{8} \mathrm{f}=0\)So, centre is \((-\mathrm{g},-\mathrm{f})=\left(-\frac{3}{8}, 0\right)\)
GUJCET-2011
Conic Section
119722
The centre and radius of the circle \(x^2+y^2-4 x\) \(+2 \mathbf{y}=0\) are
1 \((2,-1)\) and 5
2 \((4,2)\) and \(\sqrt{20}\)
3 \((2,-1)\) and \(\sqrt{5}\)
4 \((-2,1)\) and 5
5 \((-2,1)\) and \(\sqrt{5}\)
Explanation:
C Given, \(x^2+y^2-4 x+2 y=0\) Here, \(2 g=-4,2 f=2, c=0\) \(g=-2, f=1\) Then centre, \((-\mathrm{g},-\mathrm{f})=(2,-1)\) And, \(\mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(\mathrm{r}=\sqrt{(-2)^2+(1)^2-0}\) \(\mathrm{r}=\sqrt{4+1}=\sqrt{5}\)
Kerala CEE-2020
Conic Section
119647
Consider the circles \(x^2+(y-1)^2=9\), \((x-1)^2+y^2=25\). They are such that
1 these circles touch each other
2 One of these circles lies entirely inside the other
3 each of these circles lies outside the other
4 they intersect in two points
Explanation:
B The circle are \(x^2+(y-1)^2=9\) \(\text { And, }\) \((x-1)^2+(y)^2=25\) \(c_1=(0,1), c_2=(1,0)\) \(\quad r_1=3, \quad r_2=5\) \(\mathrm{c}_1 \mathrm{c}_2=\sqrt{1^2+1^2}=\sqrt{2}\) \(\mathrm{r}_2-\mathrm{r}_1=2\) \(\mathrm{c}_1 \mathrm{c}_2\lt \left(\mathrm{r}_2-\mathrm{r}_1\right)\) \(\sqrt{2}\lt \mathrm{r}_2-\mathrm{r}_1\)Therefore, one circle lies entirely inside the other circle.
