120013
Two circles with equal radii are intersecting at the points \((0,1)\) and \((0,-1)\). The tangent at the point \((0,1)\) to one of the circles passes through the centre of the other circle. Then, the distance between the centres of these circles is
1 \(\sqrt{2}\)
2 \(2 \sqrt{2}\)
3 1
4 2
Explanation:
D : \(\because\) Two circles of equal radii intersect each other orthogonally. Then, \(\mathrm{M}\) is midpoint of \(\mathrm{PQ}\). Hence, \(\mathrm{PM}=\mathrm{C}_1 \mathrm{M}=\mathrm{C}_2 \mathrm{M}\) \(\mathrm{PM}=\frac{1}{2} \sqrt{(0-0)^2+(1+1)^2}=1\) Distance between centers \(=1+1=2\) 2
JEE Main 11.01.2019
Conic Section
120015
The circle passing through the intersection of the circles, \(x^2+y^2-6 x=0\) and \(x^2+y^2-4 y=0\), having its centre on the line, \(2 x-3 y+12=0\), also passes through the point
1 \((-1,3)\)
2 \((-3,1)\)
3 \((1,-3)\)
4 \((-3,6)\)
Explanation:
D Given circle equation, \(S_1=x^2+y^2-6 x=0\) \(S_2=x^2+y^2-4 y=0\) Let \(S\) be the circle passing through point of intersection of \(\mathrm{S}_1\) and \(\mathrm{S}_2\) \(\therefore\) \(S_1+\lambda S_2=0\) \(\left(x^2+y^2-6 x\right)+\lambda\left(x^2+y^2-4 y\right)=0\) \(x^2+y^2-\left(\frac{6}{1+\lambda}\right) x-\left(\frac{4 \lambda}{1+\lambda}\right) y=0\) Centre, \(\left(\frac{3}{1+\lambda}, \frac{2 \lambda}{1+\lambda}\right)\) lies on \(2 x-3 y+12=0\) \(\frac{6}{\lambda+1}-\frac{6 \lambda}{\lambda+1}+12=0\) \(\lambda=-3\) put in eq \({ }^{\mathrm{n}}\).(i) \(\mathrm{S}=\mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}-6 \mathrm{y}=0\)Now check option point \((-3,6)\) lies on \(S\)
JEE Main 04.09.2020
Conic Section
120016
The line \(2 x-y+1=0\) is a tangent to the circle at the point \((2,5)\) and the centre of the circle lies on \(x-2 y=4\). Then, the radius of the circle is
1 \(3 \sqrt{5}\)
2 \(5 \sqrt{3}\)
3 \(5 \sqrt{4}\)
4 \(4 \sqrt{5}\)
Explanation:
A Given tangent, \(\mathrm{T}=2 \mathrm{x}-\mathrm{y}+1=0\) Slope of tangent, \(\mathrm{T}\) to the circle is 2 the line joining the centre and the point of contact of tangent is perpendicular with the tangent. Thus slope of line joining point of contact and centre is \(-\frac{1}{2}\) and thus the equation of the line is \((y-5)=-\frac{1}{2}(x-2)\) \(\mathrm{x}+2 \mathrm{y}=12\) \(\therefore\) Intersection with \(\mathrm{x}-2 \mathrm{y}=4\) will give coordinates of centre which are \((8,2)\) \(\therefore \mathrm{r}=\mathrm{OA}=\sqrt{(8-2)^2+(2-5)^2}\) \(\mathrm{r}=3 \sqrt{5}\)
JEE Main 17.03.2021
Conic Section
120017
Choose the incorrect statement about the two circles whose equations are given below \(x^2+y^2-10 x-10 y+41=0\) and \(x^2+y^2-16 x-10 y+80=0\)
1 Distance between two centres is the average of radii of both the circles.
2 Both circles' centres lie inside region of one another.
3 Both circles pass through the centre of each other.
4 Circles have two intersection points.
Explanation:
B Given circle equation, \(S_1=x^2+y^2-10 x-10 y+41=0\) \(S_2=x^2+y^2-16 x-10 y+80=0\) Compare the general equation of the circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fx}+\mathrm{c}=0\) Whose centre \((-\mathrm{g},-\mathrm{f})\) and radius \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) Circle, \(\mathrm{S}_1=\mathrm{C}_1(5,5)\) And, radius \(\mathrm{R}_1=\sqrt{5^2+5^2-41}=\sqrt{9}=3\) Now circle, \(\mathrm{S}_2=\mathrm{C}_2(8,5)\) Radius, \(\mathrm{R}_2=\sqrt{(8)^2+(5)^2-80}=\sqrt{9}=3\) The position of \(\mathrm{C}_1=(5,5)\) in circle (2) \(=25+25-80-50+80=0\) The position of \(\mathrm{C}_2=(8,5)\) in circle (1) \(64+25-80-50+41=0\)Thus both circle centers lie inside the region of one another.
120013
Two circles with equal radii are intersecting at the points \((0,1)\) and \((0,-1)\). The tangent at the point \((0,1)\) to one of the circles passes through the centre of the other circle. Then, the distance between the centres of these circles is
1 \(\sqrt{2}\)
2 \(2 \sqrt{2}\)
3 1
4 2
Explanation:
D : \(\because\) Two circles of equal radii intersect each other orthogonally. Then, \(\mathrm{M}\) is midpoint of \(\mathrm{PQ}\). Hence, \(\mathrm{PM}=\mathrm{C}_1 \mathrm{M}=\mathrm{C}_2 \mathrm{M}\) \(\mathrm{PM}=\frac{1}{2} \sqrt{(0-0)^2+(1+1)^2}=1\) Distance between centers \(=1+1=2\) 2
JEE Main 11.01.2019
Conic Section
120015
The circle passing through the intersection of the circles, \(x^2+y^2-6 x=0\) and \(x^2+y^2-4 y=0\), having its centre on the line, \(2 x-3 y+12=0\), also passes through the point
1 \((-1,3)\)
2 \((-3,1)\)
3 \((1,-3)\)
4 \((-3,6)\)
Explanation:
D Given circle equation, \(S_1=x^2+y^2-6 x=0\) \(S_2=x^2+y^2-4 y=0\) Let \(S\) be the circle passing through point of intersection of \(\mathrm{S}_1\) and \(\mathrm{S}_2\) \(\therefore\) \(S_1+\lambda S_2=0\) \(\left(x^2+y^2-6 x\right)+\lambda\left(x^2+y^2-4 y\right)=0\) \(x^2+y^2-\left(\frac{6}{1+\lambda}\right) x-\left(\frac{4 \lambda}{1+\lambda}\right) y=0\) Centre, \(\left(\frac{3}{1+\lambda}, \frac{2 \lambda}{1+\lambda}\right)\) lies on \(2 x-3 y+12=0\) \(\frac{6}{\lambda+1}-\frac{6 \lambda}{\lambda+1}+12=0\) \(\lambda=-3\) put in eq \({ }^{\mathrm{n}}\).(i) \(\mathrm{S}=\mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}-6 \mathrm{y}=0\)Now check option point \((-3,6)\) lies on \(S\)
JEE Main 04.09.2020
Conic Section
120016
The line \(2 x-y+1=0\) is a tangent to the circle at the point \((2,5)\) and the centre of the circle lies on \(x-2 y=4\). Then, the radius of the circle is
1 \(3 \sqrt{5}\)
2 \(5 \sqrt{3}\)
3 \(5 \sqrt{4}\)
4 \(4 \sqrt{5}\)
Explanation:
A Given tangent, \(\mathrm{T}=2 \mathrm{x}-\mathrm{y}+1=0\) Slope of tangent, \(\mathrm{T}\) to the circle is 2 the line joining the centre and the point of contact of tangent is perpendicular with the tangent. Thus slope of line joining point of contact and centre is \(-\frac{1}{2}\) and thus the equation of the line is \((y-5)=-\frac{1}{2}(x-2)\) \(\mathrm{x}+2 \mathrm{y}=12\) \(\therefore\) Intersection with \(\mathrm{x}-2 \mathrm{y}=4\) will give coordinates of centre which are \((8,2)\) \(\therefore \mathrm{r}=\mathrm{OA}=\sqrt{(8-2)^2+(2-5)^2}\) \(\mathrm{r}=3 \sqrt{5}\)
JEE Main 17.03.2021
Conic Section
120017
Choose the incorrect statement about the two circles whose equations are given below \(x^2+y^2-10 x-10 y+41=0\) and \(x^2+y^2-16 x-10 y+80=0\)
1 Distance between two centres is the average of radii of both the circles.
2 Both circles' centres lie inside region of one another.
3 Both circles pass through the centre of each other.
4 Circles have two intersection points.
Explanation:
B Given circle equation, \(S_1=x^2+y^2-10 x-10 y+41=0\) \(S_2=x^2+y^2-16 x-10 y+80=0\) Compare the general equation of the circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fx}+\mathrm{c}=0\) Whose centre \((-\mathrm{g},-\mathrm{f})\) and radius \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) Circle, \(\mathrm{S}_1=\mathrm{C}_1(5,5)\) And, radius \(\mathrm{R}_1=\sqrt{5^2+5^2-41}=\sqrt{9}=3\) Now circle, \(\mathrm{S}_2=\mathrm{C}_2(8,5)\) Radius, \(\mathrm{R}_2=\sqrt{(8)^2+(5)^2-80}=\sqrt{9}=3\) The position of \(\mathrm{C}_1=(5,5)\) in circle (2) \(=25+25-80-50+80=0\) The position of \(\mathrm{C}_2=(8,5)\) in circle (1) \(64+25-80-50+41=0\)Thus both circle centers lie inside the region of one another.
120013
Two circles with equal radii are intersecting at the points \((0,1)\) and \((0,-1)\). The tangent at the point \((0,1)\) to one of the circles passes through the centre of the other circle. Then, the distance between the centres of these circles is
1 \(\sqrt{2}\)
2 \(2 \sqrt{2}\)
3 1
4 2
Explanation:
D : \(\because\) Two circles of equal radii intersect each other orthogonally. Then, \(\mathrm{M}\) is midpoint of \(\mathrm{PQ}\). Hence, \(\mathrm{PM}=\mathrm{C}_1 \mathrm{M}=\mathrm{C}_2 \mathrm{M}\) \(\mathrm{PM}=\frac{1}{2} \sqrt{(0-0)^2+(1+1)^2}=1\) Distance between centers \(=1+1=2\) 2
JEE Main 11.01.2019
Conic Section
120015
The circle passing through the intersection of the circles, \(x^2+y^2-6 x=0\) and \(x^2+y^2-4 y=0\), having its centre on the line, \(2 x-3 y+12=0\), also passes through the point
1 \((-1,3)\)
2 \((-3,1)\)
3 \((1,-3)\)
4 \((-3,6)\)
Explanation:
D Given circle equation, \(S_1=x^2+y^2-6 x=0\) \(S_2=x^2+y^2-4 y=0\) Let \(S\) be the circle passing through point of intersection of \(\mathrm{S}_1\) and \(\mathrm{S}_2\) \(\therefore\) \(S_1+\lambda S_2=0\) \(\left(x^2+y^2-6 x\right)+\lambda\left(x^2+y^2-4 y\right)=0\) \(x^2+y^2-\left(\frac{6}{1+\lambda}\right) x-\left(\frac{4 \lambda}{1+\lambda}\right) y=0\) Centre, \(\left(\frac{3}{1+\lambda}, \frac{2 \lambda}{1+\lambda}\right)\) lies on \(2 x-3 y+12=0\) \(\frac{6}{\lambda+1}-\frac{6 \lambda}{\lambda+1}+12=0\) \(\lambda=-3\) put in eq \({ }^{\mathrm{n}}\).(i) \(\mathrm{S}=\mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}-6 \mathrm{y}=0\)Now check option point \((-3,6)\) lies on \(S\)
JEE Main 04.09.2020
Conic Section
120016
The line \(2 x-y+1=0\) is a tangent to the circle at the point \((2,5)\) and the centre of the circle lies on \(x-2 y=4\). Then, the radius of the circle is
1 \(3 \sqrt{5}\)
2 \(5 \sqrt{3}\)
3 \(5 \sqrt{4}\)
4 \(4 \sqrt{5}\)
Explanation:
A Given tangent, \(\mathrm{T}=2 \mathrm{x}-\mathrm{y}+1=0\) Slope of tangent, \(\mathrm{T}\) to the circle is 2 the line joining the centre and the point of contact of tangent is perpendicular with the tangent. Thus slope of line joining point of contact and centre is \(-\frac{1}{2}\) and thus the equation of the line is \((y-5)=-\frac{1}{2}(x-2)\) \(\mathrm{x}+2 \mathrm{y}=12\) \(\therefore\) Intersection with \(\mathrm{x}-2 \mathrm{y}=4\) will give coordinates of centre which are \((8,2)\) \(\therefore \mathrm{r}=\mathrm{OA}=\sqrt{(8-2)^2+(2-5)^2}\) \(\mathrm{r}=3 \sqrt{5}\)
JEE Main 17.03.2021
Conic Section
120017
Choose the incorrect statement about the two circles whose equations are given below \(x^2+y^2-10 x-10 y+41=0\) and \(x^2+y^2-16 x-10 y+80=0\)
1 Distance between two centres is the average of radii of both the circles.
2 Both circles' centres lie inside region of one another.
3 Both circles pass through the centre of each other.
4 Circles have two intersection points.
Explanation:
B Given circle equation, \(S_1=x^2+y^2-10 x-10 y+41=0\) \(S_2=x^2+y^2-16 x-10 y+80=0\) Compare the general equation of the circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fx}+\mathrm{c}=0\) Whose centre \((-\mathrm{g},-\mathrm{f})\) and radius \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) Circle, \(\mathrm{S}_1=\mathrm{C}_1(5,5)\) And, radius \(\mathrm{R}_1=\sqrt{5^2+5^2-41}=\sqrt{9}=3\) Now circle, \(\mathrm{S}_2=\mathrm{C}_2(8,5)\) Radius, \(\mathrm{R}_2=\sqrt{(8)^2+(5)^2-80}=\sqrt{9}=3\) The position of \(\mathrm{C}_1=(5,5)\) in circle (2) \(=25+25-80-50+80=0\) The position of \(\mathrm{C}_2=(8,5)\) in circle (1) \(64+25-80-50+41=0\)Thus both circle centers lie inside the region of one another.
120013
Two circles with equal radii are intersecting at the points \((0,1)\) and \((0,-1)\). The tangent at the point \((0,1)\) to one of the circles passes through the centre of the other circle. Then, the distance between the centres of these circles is
1 \(\sqrt{2}\)
2 \(2 \sqrt{2}\)
3 1
4 2
Explanation:
D : \(\because\) Two circles of equal radii intersect each other orthogonally. Then, \(\mathrm{M}\) is midpoint of \(\mathrm{PQ}\). Hence, \(\mathrm{PM}=\mathrm{C}_1 \mathrm{M}=\mathrm{C}_2 \mathrm{M}\) \(\mathrm{PM}=\frac{1}{2} \sqrt{(0-0)^2+(1+1)^2}=1\) Distance between centers \(=1+1=2\) 2
JEE Main 11.01.2019
Conic Section
120015
The circle passing through the intersection of the circles, \(x^2+y^2-6 x=0\) and \(x^2+y^2-4 y=0\), having its centre on the line, \(2 x-3 y+12=0\), also passes through the point
1 \((-1,3)\)
2 \((-3,1)\)
3 \((1,-3)\)
4 \((-3,6)\)
Explanation:
D Given circle equation, \(S_1=x^2+y^2-6 x=0\) \(S_2=x^2+y^2-4 y=0\) Let \(S\) be the circle passing through point of intersection of \(\mathrm{S}_1\) and \(\mathrm{S}_2\) \(\therefore\) \(S_1+\lambda S_2=0\) \(\left(x^2+y^2-6 x\right)+\lambda\left(x^2+y^2-4 y\right)=0\) \(x^2+y^2-\left(\frac{6}{1+\lambda}\right) x-\left(\frac{4 \lambda}{1+\lambda}\right) y=0\) Centre, \(\left(\frac{3}{1+\lambda}, \frac{2 \lambda}{1+\lambda}\right)\) lies on \(2 x-3 y+12=0\) \(\frac{6}{\lambda+1}-\frac{6 \lambda}{\lambda+1}+12=0\) \(\lambda=-3\) put in eq \({ }^{\mathrm{n}}\).(i) \(\mathrm{S}=\mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}-6 \mathrm{y}=0\)Now check option point \((-3,6)\) lies on \(S\)
JEE Main 04.09.2020
Conic Section
120016
The line \(2 x-y+1=0\) is a tangent to the circle at the point \((2,5)\) and the centre of the circle lies on \(x-2 y=4\). Then, the radius of the circle is
1 \(3 \sqrt{5}\)
2 \(5 \sqrt{3}\)
3 \(5 \sqrt{4}\)
4 \(4 \sqrt{5}\)
Explanation:
A Given tangent, \(\mathrm{T}=2 \mathrm{x}-\mathrm{y}+1=0\) Slope of tangent, \(\mathrm{T}\) to the circle is 2 the line joining the centre and the point of contact of tangent is perpendicular with the tangent. Thus slope of line joining point of contact and centre is \(-\frac{1}{2}\) and thus the equation of the line is \((y-5)=-\frac{1}{2}(x-2)\) \(\mathrm{x}+2 \mathrm{y}=12\) \(\therefore\) Intersection with \(\mathrm{x}-2 \mathrm{y}=4\) will give coordinates of centre which are \((8,2)\) \(\therefore \mathrm{r}=\mathrm{OA}=\sqrt{(8-2)^2+(2-5)^2}\) \(\mathrm{r}=3 \sqrt{5}\)
JEE Main 17.03.2021
Conic Section
120017
Choose the incorrect statement about the two circles whose equations are given below \(x^2+y^2-10 x-10 y+41=0\) and \(x^2+y^2-16 x-10 y+80=0\)
1 Distance between two centres is the average of radii of both the circles.
2 Both circles' centres lie inside region of one another.
3 Both circles pass through the centre of each other.
4 Circles have two intersection points.
Explanation:
B Given circle equation, \(S_1=x^2+y^2-10 x-10 y+41=0\) \(S_2=x^2+y^2-16 x-10 y+80=0\) Compare the general equation of the circle \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fx}+\mathrm{c}=0\) Whose centre \((-\mathrm{g},-\mathrm{f})\) and radius \(=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) Circle, \(\mathrm{S}_1=\mathrm{C}_1(5,5)\) And, radius \(\mathrm{R}_1=\sqrt{5^2+5^2-41}=\sqrt{9}=3\) Now circle, \(\mathrm{S}_2=\mathrm{C}_2(8,5)\) Radius, \(\mathrm{R}_2=\sqrt{(8)^2+(5)^2-80}=\sqrt{9}=3\) The position of \(\mathrm{C}_1=(5,5)\) in circle (2) \(=25+25-80-50+80=0\) The position of \(\mathrm{C}_2=(8,5)\) in circle (1) \(64+25-80-50+41=0\)Thus both circle centers lie inside the region of one another.
