119642
The equation of the circle with origin as centre passing the vertices of an equilateral triangle whose median is of length \(3 a\) is
1 \(x^2+y^2=9 a^2\)
2 \(x^2+y^2=16 a^2\)
3 \(x^2+y^2=a^2\)
4 None of the above
Explanation:
D : A median of a triangle is a line segment joining a vertex to the mid-point of the opposite side thus bisecting that side. Let, \(\mathrm{ABC}\) be an equilateral triangle in which median \(\mathrm{AD}=3 \mathrm{a}\) Centre of the circle is same as the centroid of the triangle i.e. \((0,0)\) \(\mathrm{AG}: \mathrm{GD}=2: 1\) \(\mathrm{AG} =\frac{2}{3} \mathrm{AD}\) \(=\frac{2}{3} \times 3 \mathrm{a}=2 \mathrm{a}\) \(\mathrm{AG}=2 \mathrm{a}\). \(\therefore\) The equation of the circle is \((x-0)^2+(y-0)^2=(2 a)^2\) \(x^2+y^2=4 a^2\)
UPSEE-2012
Conic Section
119643
The equation of a circle with centre at \((2,2)\) and passes through the point \((4,5)\) is
1 \(x^2+y^2-4 x-4 y-5=0\)
2 \(x^2+y^2+4 x+4 y-5=0\)
3 \(x^2+y^2=5\)
4 None of these
Explanation:
A : Given data, Centre of circle \(=(2,2)\) According to question circle passes through \((4,5)\) So the point is \((2,2) \&(4,5)\) As we know that \(\mathrm{r}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\) \(r=\sqrt{(4-2)^2+(5-2)^2}\) \(\mathrm{r}=\sqrt{4+9}\) \(\mathrm{r}=\sqrt{13}\) Now equation of the circle is given by \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\) where \(\mathrm{h}\) and \(\mathrm{k}=2\), and 2 . \((x-2)^2+(y-2)^2=(\sqrt{13})^2\) \(x^2+4-4 x+y^2+4-4 y=13\) \(x^2+y^2-4 y-4 x-5=0\)
JCECE-2019]
Conic Section
119644
The equation of a circle passing through the point \((1,1)\) and the point of intersection of the circles \(x^2+y^2+13 x-3 y=0\) and \(2 x^2+2 y^2+4 x-7 y-25=0\) is
1 \(4 x^2+4 y^2+30 x-13 y-25=0\)
2 \(4 x^2+4 y^2+30 x-13 y+25=0\)
3 \(4 x^2-4 y^2-30 x+13 y-25=0\)
4 \(4 x^2-4 y^2+30 x-13 y-25=0\)
Explanation:
A Given, Circle \(-s_1 \equiv x^2+y^2+13 x-3 y=0\) Circle \(-\mathrm{s}_2 \equiv 2 \mathrm{x}^2+2 \mathrm{y}^2+4 \mathrm{x}-7 \mathrm{y}-25=0\) Concept use, \(\quad \mathrm{s}_1+\lambda \mathrm{s}_2=0\) \(\left(\mathrm{x}^2+\mathrm{y}^2+13 \mathrm{x}-3 \mathrm{y}\right)+\lambda\left(2 \mathrm{x}^2+2 \mathrm{y}^2+4 \mathrm{x}-7 \mathrm{y}-25\right)=0\) \(\cdots \text { (iii) }\) \(\because \text { The circle } \mathrm{s}_1+\lambda \mathrm{s}_2=0 \text { also passes through }(1,1)\) \(\text { So, }(1,1) \text { will satisfy the equation (iii) }\) \((1+1+13-3)+\lambda(2+2+4-7-25)=0\) \(12-24 \lambda=0\) \(\lambda=1 / 2\) Put value of \(\lambda\) is equation (iii), We get- \(\mathrm{s}_1+\lambda \mathrm{s}_2=4 \mathrm{x}^2+4 \mathrm{y}^2+30 \mathrm{x}-13 \mathrm{y}-25=0\)
JCECE-2017
Conic Section
119645
If the line \(y=\sqrt{3} x+k\) touches the circle \(x^2+y^2\) \(=16\), then the value of \(k\) is
1 \(\pm 8\)
2 \(\pm 6\)
3 \(\pm 4\)
4 \(\pm 10\)
Explanation:
A Given, circle \(x^2+y^2=16\) Here centre is \((0,0)\) and its radius is 4 . The given line is \(y=\sqrt{3} \mathrm{x}+\mathrm{k}\) According to equation, the line will touch the circle if the perpendicular distance from \(\mathrm{C}(0,0)\) to the line \(=\) radius of circle. \(\frac{|\sqrt{3} \times 0-0+\mathrm{k}|}{\sqrt{(\sqrt{3})^2+1^2}}=4\) \(\frac{|\mathrm{k}|}{2}=4\) \(\mathrm{k}= \pm 8\)
119642
The equation of the circle with origin as centre passing the vertices of an equilateral triangle whose median is of length \(3 a\) is
1 \(x^2+y^2=9 a^2\)
2 \(x^2+y^2=16 a^2\)
3 \(x^2+y^2=a^2\)
4 None of the above
Explanation:
D : A median of a triangle is a line segment joining a vertex to the mid-point of the opposite side thus bisecting that side. Let, \(\mathrm{ABC}\) be an equilateral triangle in which median \(\mathrm{AD}=3 \mathrm{a}\) Centre of the circle is same as the centroid of the triangle i.e. \((0,0)\) \(\mathrm{AG}: \mathrm{GD}=2: 1\) \(\mathrm{AG} =\frac{2}{3} \mathrm{AD}\) \(=\frac{2}{3} \times 3 \mathrm{a}=2 \mathrm{a}\) \(\mathrm{AG}=2 \mathrm{a}\). \(\therefore\) The equation of the circle is \((x-0)^2+(y-0)^2=(2 a)^2\) \(x^2+y^2=4 a^2\)
UPSEE-2012
Conic Section
119643
The equation of a circle with centre at \((2,2)\) and passes through the point \((4,5)\) is
1 \(x^2+y^2-4 x-4 y-5=0\)
2 \(x^2+y^2+4 x+4 y-5=0\)
3 \(x^2+y^2=5\)
4 None of these
Explanation:
A : Given data, Centre of circle \(=(2,2)\) According to question circle passes through \((4,5)\) So the point is \((2,2) \&(4,5)\) As we know that \(\mathrm{r}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\) \(r=\sqrt{(4-2)^2+(5-2)^2}\) \(\mathrm{r}=\sqrt{4+9}\) \(\mathrm{r}=\sqrt{13}\) Now equation of the circle is given by \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\) where \(\mathrm{h}\) and \(\mathrm{k}=2\), and 2 . \((x-2)^2+(y-2)^2=(\sqrt{13})^2\) \(x^2+4-4 x+y^2+4-4 y=13\) \(x^2+y^2-4 y-4 x-5=0\)
JCECE-2019]
Conic Section
119644
The equation of a circle passing through the point \((1,1)\) and the point of intersection of the circles \(x^2+y^2+13 x-3 y=0\) and \(2 x^2+2 y^2+4 x-7 y-25=0\) is
1 \(4 x^2+4 y^2+30 x-13 y-25=0\)
2 \(4 x^2+4 y^2+30 x-13 y+25=0\)
3 \(4 x^2-4 y^2-30 x+13 y-25=0\)
4 \(4 x^2-4 y^2+30 x-13 y-25=0\)
Explanation:
A Given, Circle \(-s_1 \equiv x^2+y^2+13 x-3 y=0\) Circle \(-\mathrm{s}_2 \equiv 2 \mathrm{x}^2+2 \mathrm{y}^2+4 \mathrm{x}-7 \mathrm{y}-25=0\) Concept use, \(\quad \mathrm{s}_1+\lambda \mathrm{s}_2=0\) \(\left(\mathrm{x}^2+\mathrm{y}^2+13 \mathrm{x}-3 \mathrm{y}\right)+\lambda\left(2 \mathrm{x}^2+2 \mathrm{y}^2+4 \mathrm{x}-7 \mathrm{y}-25\right)=0\) \(\cdots \text { (iii) }\) \(\because \text { The circle } \mathrm{s}_1+\lambda \mathrm{s}_2=0 \text { also passes through }(1,1)\) \(\text { So, }(1,1) \text { will satisfy the equation (iii) }\) \((1+1+13-3)+\lambda(2+2+4-7-25)=0\) \(12-24 \lambda=0\) \(\lambda=1 / 2\) Put value of \(\lambda\) is equation (iii), We get- \(\mathrm{s}_1+\lambda \mathrm{s}_2=4 \mathrm{x}^2+4 \mathrm{y}^2+30 \mathrm{x}-13 \mathrm{y}-25=0\)
JCECE-2017
Conic Section
119645
If the line \(y=\sqrt{3} x+k\) touches the circle \(x^2+y^2\) \(=16\), then the value of \(k\) is
1 \(\pm 8\)
2 \(\pm 6\)
3 \(\pm 4\)
4 \(\pm 10\)
Explanation:
A Given, circle \(x^2+y^2=16\) Here centre is \((0,0)\) and its radius is 4 . The given line is \(y=\sqrt{3} \mathrm{x}+\mathrm{k}\) According to equation, the line will touch the circle if the perpendicular distance from \(\mathrm{C}(0,0)\) to the line \(=\) radius of circle. \(\frac{|\sqrt{3} \times 0-0+\mathrm{k}|}{\sqrt{(\sqrt{3})^2+1^2}}=4\) \(\frac{|\mathrm{k}|}{2}=4\) \(\mathrm{k}= \pm 8\)
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Conic Section
119642
The equation of the circle with origin as centre passing the vertices of an equilateral triangle whose median is of length \(3 a\) is
1 \(x^2+y^2=9 a^2\)
2 \(x^2+y^2=16 a^2\)
3 \(x^2+y^2=a^2\)
4 None of the above
Explanation:
D : A median of a triangle is a line segment joining a vertex to the mid-point of the opposite side thus bisecting that side. Let, \(\mathrm{ABC}\) be an equilateral triangle in which median \(\mathrm{AD}=3 \mathrm{a}\) Centre of the circle is same as the centroid of the triangle i.e. \((0,0)\) \(\mathrm{AG}: \mathrm{GD}=2: 1\) \(\mathrm{AG} =\frac{2}{3} \mathrm{AD}\) \(=\frac{2}{3} \times 3 \mathrm{a}=2 \mathrm{a}\) \(\mathrm{AG}=2 \mathrm{a}\). \(\therefore\) The equation of the circle is \((x-0)^2+(y-0)^2=(2 a)^2\) \(x^2+y^2=4 a^2\)
UPSEE-2012
Conic Section
119643
The equation of a circle with centre at \((2,2)\) and passes through the point \((4,5)\) is
1 \(x^2+y^2-4 x-4 y-5=0\)
2 \(x^2+y^2+4 x+4 y-5=0\)
3 \(x^2+y^2=5\)
4 None of these
Explanation:
A : Given data, Centre of circle \(=(2,2)\) According to question circle passes through \((4,5)\) So the point is \((2,2) \&(4,5)\) As we know that \(\mathrm{r}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\) \(r=\sqrt{(4-2)^2+(5-2)^2}\) \(\mathrm{r}=\sqrt{4+9}\) \(\mathrm{r}=\sqrt{13}\) Now equation of the circle is given by \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\) where \(\mathrm{h}\) and \(\mathrm{k}=2\), and 2 . \((x-2)^2+(y-2)^2=(\sqrt{13})^2\) \(x^2+4-4 x+y^2+4-4 y=13\) \(x^2+y^2-4 y-4 x-5=0\)
JCECE-2019]
Conic Section
119644
The equation of a circle passing through the point \((1,1)\) and the point of intersection of the circles \(x^2+y^2+13 x-3 y=0\) and \(2 x^2+2 y^2+4 x-7 y-25=0\) is
1 \(4 x^2+4 y^2+30 x-13 y-25=0\)
2 \(4 x^2+4 y^2+30 x-13 y+25=0\)
3 \(4 x^2-4 y^2-30 x+13 y-25=0\)
4 \(4 x^2-4 y^2+30 x-13 y-25=0\)
Explanation:
A Given, Circle \(-s_1 \equiv x^2+y^2+13 x-3 y=0\) Circle \(-\mathrm{s}_2 \equiv 2 \mathrm{x}^2+2 \mathrm{y}^2+4 \mathrm{x}-7 \mathrm{y}-25=0\) Concept use, \(\quad \mathrm{s}_1+\lambda \mathrm{s}_2=0\) \(\left(\mathrm{x}^2+\mathrm{y}^2+13 \mathrm{x}-3 \mathrm{y}\right)+\lambda\left(2 \mathrm{x}^2+2 \mathrm{y}^2+4 \mathrm{x}-7 \mathrm{y}-25\right)=0\) \(\cdots \text { (iii) }\) \(\because \text { The circle } \mathrm{s}_1+\lambda \mathrm{s}_2=0 \text { also passes through }(1,1)\) \(\text { So, }(1,1) \text { will satisfy the equation (iii) }\) \((1+1+13-3)+\lambda(2+2+4-7-25)=0\) \(12-24 \lambda=0\) \(\lambda=1 / 2\) Put value of \(\lambda\) is equation (iii), We get- \(\mathrm{s}_1+\lambda \mathrm{s}_2=4 \mathrm{x}^2+4 \mathrm{y}^2+30 \mathrm{x}-13 \mathrm{y}-25=0\)
JCECE-2017
Conic Section
119645
If the line \(y=\sqrt{3} x+k\) touches the circle \(x^2+y^2\) \(=16\), then the value of \(k\) is
1 \(\pm 8\)
2 \(\pm 6\)
3 \(\pm 4\)
4 \(\pm 10\)
Explanation:
A Given, circle \(x^2+y^2=16\) Here centre is \((0,0)\) and its radius is 4 . The given line is \(y=\sqrt{3} \mathrm{x}+\mathrm{k}\) According to equation, the line will touch the circle if the perpendicular distance from \(\mathrm{C}(0,0)\) to the line \(=\) radius of circle. \(\frac{|\sqrt{3} \times 0-0+\mathrm{k}|}{\sqrt{(\sqrt{3})^2+1^2}}=4\) \(\frac{|\mathrm{k}|}{2}=4\) \(\mathrm{k}= \pm 8\)
119642
The equation of the circle with origin as centre passing the vertices of an equilateral triangle whose median is of length \(3 a\) is
1 \(x^2+y^2=9 a^2\)
2 \(x^2+y^2=16 a^2\)
3 \(x^2+y^2=a^2\)
4 None of the above
Explanation:
D : A median of a triangle is a line segment joining a vertex to the mid-point of the opposite side thus bisecting that side. Let, \(\mathrm{ABC}\) be an equilateral triangle in which median \(\mathrm{AD}=3 \mathrm{a}\) Centre of the circle is same as the centroid of the triangle i.e. \((0,0)\) \(\mathrm{AG}: \mathrm{GD}=2: 1\) \(\mathrm{AG} =\frac{2}{3} \mathrm{AD}\) \(=\frac{2}{3} \times 3 \mathrm{a}=2 \mathrm{a}\) \(\mathrm{AG}=2 \mathrm{a}\). \(\therefore\) The equation of the circle is \((x-0)^2+(y-0)^2=(2 a)^2\) \(x^2+y^2=4 a^2\)
UPSEE-2012
Conic Section
119643
The equation of a circle with centre at \((2,2)\) and passes through the point \((4,5)\) is
1 \(x^2+y^2-4 x-4 y-5=0\)
2 \(x^2+y^2+4 x+4 y-5=0\)
3 \(x^2+y^2=5\)
4 None of these
Explanation:
A : Given data, Centre of circle \(=(2,2)\) According to question circle passes through \((4,5)\) So the point is \((2,2) \&(4,5)\) As we know that \(\mathrm{r}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\) \(r=\sqrt{(4-2)^2+(5-2)^2}\) \(\mathrm{r}=\sqrt{4+9}\) \(\mathrm{r}=\sqrt{13}\) Now equation of the circle is given by \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\) where \(\mathrm{h}\) and \(\mathrm{k}=2\), and 2 . \((x-2)^2+(y-2)^2=(\sqrt{13})^2\) \(x^2+4-4 x+y^2+4-4 y=13\) \(x^2+y^2-4 y-4 x-5=0\)
JCECE-2019]
Conic Section
119644
The equation of a circle passing through the point \((1,1)\) and the point of intersection of the circles \(x^2+y^2+13 x-3 y=0\) and \(2 x^2+2 y^2+4 x-7 y-25=0\) is
1 \(4 x^2+4 y^2+30 x-13 y-25=0\)
2 \(4 x^2+4 y^2+30 x-13 y+25=0\)
3 \(4 x^2-4 y^2-30 x+13 y-25=0\)
4 \(4 x^2-4 y^2+30 x-13 y-25=0\)
Explanation:
A Given, Circle \(-s_1 \equiv x^2+y^2+13 x-3 y=0\) Circle \(-\mathrm{s}_2 \equiv 2 \mathrm{x}^2+2 \mathrm{y}^2+4 \mathrm{x}-7 \mathrm{y}-25=0\) Concept use, \(\quad \mathrm{s}_1+\lambda \mathrm{s}_2=0\) \(\left(\mathrm{x}^2+\mathrm{y}^2+13 \mathrm{x}-3 \mathrm{y}\right)+\lambda\left(2 \mathrm{x}^2+2 \mathrm{y}^2+4 \mathrm{x}-7 \mathrm{y}-25\right)=0\) \(\cdots \text { (iii) }\) \(\because \text { The circle } \mathrm{s}_1+\lambda \mathrm{s}_2=0 \text { also passes through }(1,1)\) \(\text { So, }(1,1) \text { will satisfy the equation (iii) }\) \((1+1+13-3)+\lambda(2+2+4-7-25)=0\) \(12-24 \lambda=0\) \(\lambda=1 / 2\) Put value of \(\lambda\) is equation (iii), We get- \(\mathrm{s}_1+\lambda \mathrm{s}_2=4 \mathrm{x}^2+4 \mathrm{y}^2+30 \mathrm{x}-13 \mathrm{y}-25=0\)
JCECE-2017
Conic Section
119645
If the line \(y=\sqrt{3} x+k\) touches the circle \(x^2+y^2\) \(=16\), then the value of \(k\) is
1 \(\pm 8\)
2 \(\pm 6\)
3 \(\pm 4\)
4 \(\pm 10\)
Explanation:
A Given, circle \(x^2+y^2=16\) Here centre is \((0,0)\) and its radius is 4 . The given line is \(y=\sqrt{3} \mathrm{x}+\mathrm{k}\) According to equation, the line will touch the circle if the perpendicular distance from \(\mathrm{C}(0,0)\) to the line \(=\) radius of circle. \(\frac{|\sqrt{3} \times 0-0+\mathrm{k}|}{\sqrt{(\sqrt{3})^2+1^2}}=4\) \(\frac{|\mathrm{k}|}{2}=4\) \(\mathrm{k}= \pm 8\)
