119740
The equation of the circle whose end points of a diameter are the centers of the circles \(x^2+y^2+2 x-4 y+1=0\) and \(x^2+y^2-8 x+6 y+17=0\) is
1 \(x^2+y^2+3 x-y-10=0\)
2 \(x^2+y^2-3 x-y-10=0\)
3 \(x^2+y^2-3 x+y-10=0\)
4 \(x^2+y^2+3 x+y-10=0\)
Explanation:
C :Given that, The equation of the circle are \(x^2+y^2+2 x-4 y+1=0\) \(x^2+y^2-8 x+6 y+17=0\) Let centers of given circles be \(A(-1,2)\) and \(B(4,-3)\) By diameter from of equation of circle, we write \((\mathrm{x}+1)(\mathrm{x}-4)+(\mathrm{y}-2)(\mathrm{y}+3)=0\) \(\therefore \quad \mathrm{x}^2-4 \mathrm{x}+\mathrm{x}-4+\mathrm{y}^2+3 \mathrm{y}-2 \mathrm{y}-6=0\) \(\therefore \quad \mathrm{x}^2+\mathrm{y}^2-3 \mathrm{x}+\mathrm{y}-10=0\)
MHT CET-2020
Conic Section
119726
\(A B C D\) is a square whose side is ' \(a\) '. The equation of a circle circumscribing the square, taking \(A B\) and \(A D\) as axes of references is
1 \(x^2+y^2+a x+a y=0\)
2 \(x^2+y^2+a x-a y=0\)
3 \(x^2+y^2-a x-a y=0\)
4 None of these
Explanation:
C :Taking \(A B\) and \(A D\) as axes of references we have - \(\mathrm{A}=(0,0), \mathrm{B}(\mathrm{a}, 0), \mathrm{D}(0, \mathrm{a}), \mathrm{C}=(\mathrm{a}, \mathrm{a})\) Let \(\mathrm{O}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the centre of the circle. \(\therefore \mathrm{x}_1=\frac{0+\mathrm{a}}{2} \quad \mathrm{y}_1=\frac{\mathrm{a}+0}{2}\) \(\mathrm{x}_1=\frac{\mathrm{a}}{2} \quad \mathrm{y}_1=\frac{\mathrm{a}}{2}\) \(\Rightarrow \left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}\right)\) \(\text { Radius of circle }=\frac{\mathrm{a}}{\sqrt{2}}\) So, the equation of circle is \(\left(x-\frac{a}{2}\right)^2+\left(y-\frac{a}{2}\right)^2=\left(\frac{a}{\sqrt{2}}\right)^2\) \(x^2+\frac{a^2}{4}-2 \frac{a x}{2}+y^2+\frac{a^2}{4}-2 y \cdot \frac{a}{2}=\frac{a^2}{2}\) \(x^2+y^2-a x-a y=0\) \(x^2+y^2-a x-a y=0\)
JEEE-2011
Conic Section
119727
Four distinct points \((2 k, 3 k),(1,0),(0,1)\) and \((0,0)\) lie on a circle for
1 all integral values of k
2 \(0\lt \mathrm{k}\lt 1\)
3 \(\mathrm{k}\lt 0\)
4 None of these
Explanation:
B The equation of the circle passing through vertices \((1,0),(0,1)\) and \((0,0)\) is - \((\mathrm{x}-1)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-1)=0\) \(\Rightarrow(\mathrm{x}-1) \mathrm{x}+\mathrm{y}(\mathrm{y}-1)=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}-\mathrm{y}=0\) Which passes through \((2 \mathrm{k}, 3 \mathrm{k})\) then :- \(4 \mathrm{k}^2+9 \mathrm{k}^2-2 \mathrm{k}-3 \mathrm{k}=0\) \(\Rightarrow 13 \mathrm{k}^2-5 \mathrm{k}=0\) \(\Rightarrow \mathrm{k}(13 \mathrm{k}-5)=0\) as \(\mathrm{k} \neq 0 \therefore \mathrm{k}=\frac{5}{13} \quad(\because \mathrm{k} \neq 0\) as po int s are distinct \()\) \(\quad \therefore 0\lt \mathrm{k}\lt 1\)
JEEE-2010]
Conic Section
119728
One of the diameters of the circle \(x^2+y^2-12 x\) \(+4 y+6=0\) is
1 \(x+y=0\)
2 \(x+3 y=0\)
3 \(x=y\)
4 None of these
Explanation:
B Given circle is :- \(\mathrm{x}^2+\mathrm{y}^2-12 \mathrm{x}+4 \mathrm{y}+6=0\) centre \((\mathrm{c})=(6,-2)\) We know that, centre of the circle lies on the diameter of the circle. Out of all options - Point \((6,-2)\) satisfy the equation \(\mathrm{x}+3 \mathrm{y}=0\) i.e. \(6+3(-2)=0\) \(0=0\)
JEEE-2010
Conic Section
119729
Equation of circle with centre \((-a,-b)\) and radius \(\sqrt{a^2-b^2}\) is
1 \(x^2+y^2+2 a x+2 b y+2 b^2=0\)
2 \(x^2+y^2-2 a x-2 b y-2 b^2=0\)
3 \(x^2+y^2-2 a x-2 b y+2 b^2=0\)
4 \(x^2+y^2-2 a x+2 b y+2 a^2=0\)
Explanation:
A Given that, Center \(=(-a,-b)\) Radius \(=\sqrt{a^2-b^2}\) We know that, general equation of a circle is - \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\) Where, \((\mathrm{h}, \mathrm{k})\) centre and \(\mathrm{r}=\) radius \((x+a)^2+(y+b)^2=a^2-b^2\) \(x^2+a^2+2 a x+y^2+b^2+2 y b=a^2-b^2\) \(x^2+y^2+2 a x+2 y b+2 b^2=0 .\)
119740
The equation of the circle whose end points of a diameter are the centers of the circles \(x^2+y^2+2 x-4 y+1=0\) and \(x^2+y^2-8 x+6 y+17=0\) is
1 \(x^2+y^2+3 x-y-10=0\)
2 \(x^2+y^2-3 x-y-10=0\)
3 \(x^2+y^2-3 x+y-10=0\)
4 \(x^2+y^2+3 x+y-10=0\)
Explanation:
C :Given that, The equation of the circle are \(x^2+y^2+2 x-4 y+1=0\) \(x^2+y^2-8 x+6 y+17=0\) Let centers of given circles be \(A(-1,2)\) and \(B(4,-3)\) By diameter from of equation of circle, we write \((\mathrm{x}+1)(\mathrm{x}-4)+(\mathrm{y}-2)(\mathrm{y}+3)=0\) \(\therefore \quad \mathrm{x}^2-4 \mathrm{x}+\mathrm{x}-4+\mathrm{y}^2+3 \mathrm{y}-2 \mathrm{y}-6=0\) \(\therefore \quad \mathrm{x}^2+\mathrm{y}^2-3 \mathrm{x}+\mathrm{y}-10=0\)
MHT CET-2020
Conic Section
119726
\(A B C D\) is a square whose side is ' \(a\) '. The equation of a circle circumscribing the square, taking \(A B\) and \(A D\) as axes of references is
1 \(x^2+y^2+a x+a y=0\)
2 \(x^2+y^2+a x-a y=0\)
3 \(x^2+y^2-a x-a y=0\)
4 None of these
Explanation:
C :Taking \(A B\) and \(A D\) as axes of references we have - \(\mathrm{A}=(0,0), \mathrm{B}(\mathrm{a}, 0), \mathrm{D}(0, \mathrm{a}), \mathrm{C}=(\mathrm{a}, \mathrm{a})\) Let \(\mathrm{O}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the centre of the circle. \(\therefore \mathrm{x}_1=\frac{0+\mathrm{a}}{2} \quad \mathrm{y}_1=\frac{\mathrm{a}+0}{2}\) \(\mathrm{x}_1=\frac{\mathrm{a}}{2} \quad \mathrm{y}_1=\frac{\mathrm{a}}{2}\) \(\Rightarrow \left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}\right)\) \(\text { Radius of circle }=\frac{\mathrm{a}}{\sqrt{2}}\) So, the equation of circle is \(\left(x-\frac{a}{2}\right)^2+\left(y-\frac{a}{2}\right)^2=\left(\frac{a}{\sqrt{2}}\right)^2\) \(x^2+\frac{a^2}{4}-2 \frac{a x}{2}+y^2+\frac{a^2}{4}-2 y \cdot \frac{a}{2}=\frac{a^2}{2}\) \(x^2+y^2-a x-a y=0\) \(x^2+y^2-a x-a y=0\)
JEEE-2011
Conic Section
119727
Four distinct points \((2 k, 3 k),(1,0),(0,1)\) and \((0,0)\) lie on a circle for
1 all integral values of k
2 \(0\lt \mathrm{k}\lt 1\)
3 \(\mathrm{k}\lt 0\)
4 None of these
Explanation:
B The equation of the circle passing through vertices \((1,0),(0,1)\) and \((0,0)\) is - \((\mathrm{x}-1)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-1)=0\) \(\Rightarrow(\mathrm{x}-1) \mathrm{x}+\mathrm{y}(\mathrm{y}-1)=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}-\mathrm{y}=0\) Which passes through \((2 \mathrm{k}, 3 \mathrm{k})\) then :- \(4 \mathrm{k}^2+9 \mathrm{k}^2-2 \mathrm{k}-3 \mathrm{k}=0\) \(\Rightarrow 13 \mathrm{k}^2-5 \mathrm{k}=0\) \(\Rightarrow \mathrm{k}(13 \mathrm{k}-5)=0\) as \(\mathrm{k} \neq 0 \therefore \mathrm{k}=\frac{5}{13} \quad(\because \mathrm{k} \neq 0\) as po int s are distinct \()\) \(\quad \therefore 0\lt \mathrm{k}\lt 1\)
JEEE-2010]
Conic Section
119728
One of the diameters of the circle \(x^2+y^2-12 x\) \(+4 y+6=0\) is
1 \(x+y=0\)
2 \(x+3 y=0\)
3 \(x=y\)
4 None of these
Explanation:
B Given circle is :- \(\mathrm{x}^2+\mathrm{y}^2-12 \mathrm{x}+4 \mathrm{y}+6=0\) centre \((\mathrm{c})=(6,-2)\) We know that, centre of the circle lies on the diameter of the circle. Out of all options - Point \((6,-2)\) satisfy the equation \(\mathrm{x}+3 \mathrm{y}=0\) i.e. \(6+3(-2)=0\) \(0=0\)
JEEE-2010
Conic Section
119729
Equation of circle with centre \((-a,-b)\) and radius \(\sqrt{a^2-b^2}\) is
1 \(x^2+y^2+2 a x+2 b y+2 b^2=0\)
2 \(x^2+y^2-2 a x-2 b y-2 b^2=0\)
3 \(x^2+y^2-2 a x-2 b y+2 b^2=0\)
4 \(x^2+y^2-2 a x+2 b y+2 a^2=0\)
Explanation:
A Given that, Center \(=(-a,-b)\) Radius \(=\sqrt{a^2-b^2}\) We know that, general equation of a circle is - \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\) Where, \((\mathrm{h}, \mathrm{k})\) centre and \(\mathrm{r}=\) radius \((x+a)^2+(y+b)^2=a^2-b^2\) \(x^2+a^2+2 a x+y^2+b^2+2 y b=a^2-b^2\) \(x^2+y^2+2 a x+2 y b+2 b^2=0 .\)
119740
The equation of the circle whose end points of a diameter are the centers of the circles \(x^2+y^2+2 x-4 y+1=0\) and \(x^2+y^2-8 x+6 y+17=0\) is
1 \(x^2+y^2+3 x-y-10=0\)
2 \(x^2+y^2-3 x-y-10=0\)
3 \(x^2+y^2-3 x+y-10=0\)
4 \(x^2+y^2+3 x+y-10=0\)
Explanation:
C :Given that, The equation of the circle are \(x^2+y^2+2 x-4 y+1=0\) \(x^2+y^2-8 x+6 y+17=0\) Let centers of given circles be \(A(-1,2)\) and \(B(4,-3)\) By diameter from of equation of circle, we write \((\mathrm{x}+1)(\mathrm{x}-4)+(\mathrm{y}-2)(\mathrm{y}+3)=0\) \(\therefore \quad \mathrm{x}^2-4 \mathrm{x}+\mathrm{x}-4+\mathrm{y}^2+3 \mathrm{y}-2 \mathrm{y}-6=0\) \(\therefore \quad \mathrm{x}^2+\mathrm{y}^2-3 \mathrm{x}+\mathrm{y}-10=0\)
MHT CET-2020
Conic Section
119726
\(A B C D\) is a square whose side is ' \(a\) '. The equation of a circle circumscribing the square, taking \(A B\) and \(A D\) as axes of references is
1 \(x^2+y^2+a x+a y=0\)
2 \(x^2+y^2+a x-a y=0\)
3 \(x^2+y^2-a x-a y=0\)
4 None of these
Explanation:
C :Taking \(A B\) and \(A D\) as axes of references we have - \(\mathrm{A}=(0,0), \mathrm{B}(\mathrm{a}, 0), \mathrm{D}(0, \mathrm{a}), \mathrm{C}=(\mathrm{a}, \mathrm{a})\) Let \(\mathrm{O}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the centre of the circle. \(\therefore \mathrm{x}_1=\frac{0+\mathrm{a}}{2} \quad \mathrm{y}_1=\frac{\mathrm{a}+0}{2}\) \(\mathrm{x}_1=\frac{\mathrm{a}}{2} \quad \mathrm{y}_1=\frac{\mathrm{a}}{2}\) \(\Rightarrow \left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}\right)\) \(\text { Radius of circle }=\frac{\mathrm{a}}{\sqrt{2}}\) So, the equation of circle is \(\left(x-\frac{a}{2}\right)^2+\left(y-\frac{a}{2}\right)^2=\left(\frac{a}{\sqrt{2}}\right)^2\) \(x^2+\frac{a^2}{4}-2 \frac{a x}{2}+y^2+\frac{a^2}{4}-2 y \cdot \frac{a}{2}=\frac{a^2}{2}\) \(x^2+y^2-a x-a y=0\) \(x^2+y^2-a x-a y=0\)
JEEE-2011
Conic Section
119727
Four distinct points \((2 k, 3 k),(1,0),(0,1)\) and \((0,0)\) lie on a circle for
1 all integral values of k
2 \(0\lt \mathrm{k}\lt 1\)
3 \(\mathrm{k}\lt 0\)
4 None of these
Explanation:
B The equation of the circle passing through vertices \((1,0),(0,1)\) and \((0,0)\) is - \((\mathrm{x}-1)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-1)=0\) \(\Rightarrow(\mathrm{x}-1) \mathrm{x}+\mathrm{y}(\mathrm{y}-1)=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}-\mathrm{y}=0\) Which passes through \((2 \mathrm{k}, 3 \mathrm{k})\) then :- \(4 \mathrm{k}^2+9 \mathrm{k}^2-2 \mathrm{k}-3 \mathrm{k}=0\) \(\Rightarrow 13 \mathrm{k}^2-5 \mathrm{k}=0\) \(\Rightarrow \mathrm{k}(13 \mathrm{k}-5)=0\) as \(\mathrm{k} \neq 0 \therefore \mathrm{k}=\frac{5}{13} \quad(\because \mathrm{k} \neq 0\) as po int s are distinct \()\) \(\quad \therefore 0\lt \mathrm{k}\lt 1\)
JEEE-2010]
Conic Section
119728
One of the diameters of the circle \(x^2+y^2-12 x\) \(+4 y+6=0\) is
1 \(x+y=0\)
2 \(x+3 y=0\)
3 \(x=y\)
4 None of these
Explanation:
B Given circle is :- \(\mathrm{x}^2+\mathrm{y}^2-12 \mathrm{x}+4 \mathrm{y}+6=0\) centre \((\mathrm{c})=(6,-2)\) We know that, centre of the circle lies on the diameter of the circle. Out of all options - Point \((6,-2)\) satisfy the equation \(\mathrm{x}+3 \mathrm{y}=0\) i.e. \(6+3(-2)=0\) \(0=0\)
JEEE-2010
Conic Section
119729
Equation of circle with centre \((-a,-b)\) and radius \(\sqrt{a^2-b^2}\) is
1 \(x^2+y^2+2 a x+2 b y+2 b^2=0\)
2 \(x^2+y^2-2 a x-2 b y-2 b^2=0\)
3 \(x^2+y^2-2 a x-2 b y+2 b^2=0\)
4 \(x^2+y^2-2 a x+2 b y+2 a^2=0\)
Explanation:
A Given that, Center \(=(-a,-b)\) Radius \(=\sqrt{a^2-b^2}\) We know that, general equation of a circle is - \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\) Where, \((\mathrm{h}, \mathrm{k})\) centre and \(\mathrm{r}=\) radius \((x+a)^2+(y+b)^2=a^2-b^2\) \(x^2+a^2+2 a x+y^2+b^2+2 y b=a^2-b^2\) \(x^2+y^2+2 a x+2 y b+2 b^2=0 .\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Conic Section
119740
The equation of the circle whose end points of a diameter are the centers of the circles \(x^2+y^2+2 x-4 y+1=0\) and \(x^2+y^2-8 x+6 y+17=0\) is
1 \(x^2+y^2+3 x-y-10=0\)
2 \(x^2+y^2-3 x-y-10=0\)
3 \(x^2+y^2-3 x+y-10=0\)
4 \(x^2+y^2+3 x+y-10=0\)
Explanation:
C :Given that, The equation of the circle are \(x^2+y^2+2 x-4 y+1=0\) \(x^2+y^2-8 x+6 y+17=0\) Let centers of given circles be \(A(-1,2)\) and \(B(4,-3)\) By diameter from of equation of circle, we write \((\mathrm{x}+1)(\mathrm{x}-4)+(\mathrm{y}-2)(\mathrm{y}+3)=0\) \(\therefore \quad \mathrm{x}^2-4 \mathrm{x}+\mathrm{x}-4+\mathrm{y}^2+3 \mathrm{y}-2 \mathrm{y}-6=0\) \(\therefore \quad \mathrm{x}^2+\mathrm{y}^2-3 \mathrm{x}+\mathrm{y}-10=0\)
MHT CET-2020
Conic Section
119726
\(A B C D\) is a square whose side is ' \(a\) '. The equation of a circle circumscribing the square, taking \(A B\) and \(A D\) as axes of references is
1 \(x^2+y^2+a x+a y=0\)
2 \(x^2+y^2+a x-a y=0\)
3 \(x^2+y^2-a x-a y=0\)
4 None of these
Explanation:
C :Taking \(A B\) and \(A D\) as axes of references we have - \(\mathrm{A}=(0,0), \mathrm{B}(\mathrm{a}, 0), \mathrm{D}(0, \mathrm{a}), \mathrm{C}=(\mathrm{a}, \mathrm{a})\) Let \(\mathrm{O}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the centre of the circle. \(\therefore \mathrm{x}_1=\frac{0+\mathrm{a}}{2} \quad \mathrm{y}_1=\frac{\mathrm{a}+0}{2}\) \(\mathrm{x}_1=\frac{\mathrm{a}}{2} \quad \mathrm{y}_1=\frac{\mathrm{a}}{2}\) \(\Rightarrow \left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}\right)\) \(\text { Radius of circle }=\frac{\mathrm{a}}{\sqrt{2}}\) So, the equation of circle is \(\left(x-\frac{a}{2}\right)^2+\left(y-\frac{a}{2}\right)^2=\left(\frac{a}{\sqrt{2}}\right)^2\) \(x^2+\frac{a^2}{4}-2 \frac{a x}{2}+y^2+\frac{a^2}{4}-2 y \cdot \frac{a}{2}=\frac{a^2}{2}\) \(x^2+y^2-a x-a y=0\) \(x^2+y^2-a x-a y=0\)
JEEE-2011
Conic Section
119727
Four distinct points \((2 k, 3 k),(1,0),(0,1)\) and \((0,0)\) lie on a circle for
1 all integral values of k
2 \(0\lt \mathrm{k}\lt 1\)
3 \(\mathrm{k}\lt 0\)
4 None of these
Explanation:
B The equation of the circle passing through vertices \((1,0),(0,1)\) and \((0,0)\) is - \((\mathrm{x}-1)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-1)=0\) \(\Rightarrow(\mathrm{x}-1) \mathrm{x}+\mathrm{y}(\mathrm{y}-1)=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}-\mathrm{y}=0\) Which passes through \((2 \mathrm{k}, 3 \mathrm{k})\) then :- \(4 \mathrm{k}^2+9 \mathrm{k}^2-2 \mathrm{k}-3 \mathrm{k}=0\) \(\Rightarrow 13 \mathrm{k}^2-5 \mathrm{k}=0\) \(\Rightarrow \mathrm{k}(13 \mathrm{k}-5)=0\) as \(\mathrm{k} \neq 0 \therefore \mathrm{k}=\frac{5}{13} \quad(\because \mathrm{k} \neq 0\) as po int s are distinct \()\) \(\quad \therefore 0\lt \mathrm{k}\lt 1\)
JEEE-2010]
Conic Section
119728
One of the diameters of the circle \(x^2+y^2-12 x\) \(+4 y+6=0\) is
1 \(x+y=0\)
2 \(x+3 y=0\)
3 \(x=y\)
4 None of these
Explanation:
B Given circle is :- \(\mathrm{x}^2+\mathrm{y}^2-12 \mathrm{x}+4 \mathrm{y}+6=0\) centre \((\mathrm{c})=(6,-2)\) We know that, centre of the circle lies on the diameter of the circle. Out of all options - Point \((6,-2)\) satisfy the equation \(\mathrm{x}+3 \mathrm{y}=0\) i.e. \(6+3(-2)=0\) \(0=0\)
JEEE-2010
Conic Section
119729
Equation of circle with centre \((-a,-b)\) and radius \(\sqrt{a^2-b^2}\) is
1 \(x^2+y^2+2 a x+2 b y+2 b^2=0\)
2 \(x^2+y^2-2 a x-2 b y-2 b^2=0\)
3 \(x^2+y^2-2 a x-2 b y+2 b^2=0\)
4 \(x^2+y^2-2 a x+2 b y+2 a^2=0\)
Explanation:
A Given that, Center \(=(-a,-b)\) Radius \(=\sqrt{a^2-b^2}\) We know that, general equation of a circle is - \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\) Where, \((\mathrm{h}, \mathrm{k})\) centre and \(\mathrm{r}=\) radius \((x+a)^2+(y+b)^2=a^2-b^2\) \(x^2+a^2+2 a x+y^2+b^2+2 y b=a^2-b^2\) \(x^2+y^2+2 a x+2 y b+2 b^2=0 .\)
119740
The equation of the circle whose end points of a diameter are the centers of the circles \(x^2+y^2+2 x-4 y+1=0\) and \(x^2+y^2-8 x+6 y+17=0\) is
1 \(x^2+y^2+3 x-y-10=0\)
2 \(x^2+y^2-3 x-y-10=0\)
3 \(x^2+y^2-3 x+y-10=0\)
4 \(x^2+y^2+3 x+y-10=0\)
Explanation:
C :Given that, The equation of the circle are \(x^2+y^2+2 x-4 y+1=0\) \(x^2+y^2-8 x+6 y+17=0\) Let centers of given circles be \(A(-1,2)\) and \(B(4,-3)\) By diameter from of equation of circle, we write \((\mathrm{x}+1)(\mathrm{x}-4)+(\mathrm{y}-2)(\mathrm{y}+3)=0\) \(\therefore \quad \mathrm{x}^2-4 \mathrm{x}+\mathrm{x}-4+\mathrm{y}^2+3 \mathrm{y}-2 \mathrm{y}-6=0\) \(\therefore \quad \mathrm{x}^2+\mathrm{y}^2-3 \mathrm{x}+\mathrm{y}-10=0\)
MHT CET-2020
Conic Section
119726
\(A B C D\) is a square whose side is ' \(a\) '. The equation of a circle circumscribing the square, taking \(A B\) and \(A D\) as axes of references is
1 \(x^2+y^2+a x+a y=0\)
2 \(x^2+y^2+a x-a y=0\)
3 \(x^2+y^2-a x-a y=0\)
4 None of these
Explanation:
C :Taking \(A B\) and \(A D\) as axes of references we have - \(\mathrm{A}=(0,0), \mathrm{B}(\mathrm{a}, 0), \mathrm{D}(0, \mathrm{a}), \mathrm{C}=(\mathrm{a}, \mathrm{a})\) Let \(\mathrm{O}\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the centre of the circle. \(\therefore \mathrm{x}_1=\frac{0+\mathrm{a}}{2} \quad \mathrm{y}_1=\frac{\mathrm{a}+0}{2}\) \(\mathrm{x}_1=\frac{\mathrm{a}}{2} \quad \mathrm{y}_1=\frac{\mathrm{a}}{2}\) \(\Rightarrow \left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}\right)\) \(\text { Radius of circle }=\frac{\mathrm{a}}{\sqrt{2}}\) So, the equation of circle is \(\left(x-\frac{a}{2}\right)^2+\left(y-\frac{a}{2}\right)^2=\left(\frac{a}{\sqrt{2}}\right)^2\) \(x^2+\frac{a^2}{4}-2 \frac{a x}{2}+y^2+\frac{a^2}{4}-2 y \cdot \frac{a}{2}=\frac{a^2}{2}\) \(x^2+y^2-a x-a y=0\) \(x^2+y^2-a x-a y=0\)
JEEE-2011
Conic Section
119727
Four distinct points \((2 k, 3 k),(1,0),(0,1)\) and \((0,0)\) lie on a circle for
1 all integral values of k
2 \(0\lt \mathrm{k}\lt 1\)
3 \(\mathrm{k}\lt 0\)
4 None of these
Explanation:
B The equation of the circle passing through vertices \((1,0),(0,1)\) and \((0,0)\) is - \((\mathrm{x}-1)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-1)=0\) \(\Rightarrow(\mathrm{x}-1) \mathrm{x}+\mathrm{y}(\mathrm{y}-1)=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}-\mathrm{y}=0\) Which passes through \((2 \mathrm{k}, 3 \mathrm{k})\) then :- \(4 \mathrm{k}^2+9 \mathrm{k}^2-2 \mathrm{k}-3 \mathrm{k}=0\) \(\Rightarrow 13 \mathrm{k}^2-5 \mathrm{k}=0\) \(\Rightarrow \mathrm{k}(13 \mathrm{k}-5)=0\) as \(\mathrm{k} \neq 0 \therefore \mathrm{k}=\frac{5}{13} \quad(\because \mathrm{k} \neq 0\) as po int s are distinct \()\) \(\quad \therefore 0\lt \mathrm{k}\lt 1\)
JEEE-2010]
Conic Section
119728
One of the diameters of the circle \(x^2+y^2-12 x\) \(+4 y+6=0\) is
1 \(x+y=0\)
2 \(x+3 y=0\)
3 \(x=y\)
4 None of these
Explanation:
B Given circle is :- \(\mathrm{x}^2+\mathrm{y}^2-12 \mathrm{x}+4 \mathrm{y}+6=0\) centre \((\mathrm{c})=(6,-2)\) We know that, centre of the circle lies on the diameter of the circle. Out of all options - Point \((6,-2)\) satisfy the equation \(\mathrm{x}+3 \mathrm{y}=0\) i.e. \(6+3(-2)=0\) \(0=0\)
JEEE-2010
Conic Section
119729
Equation of circle with centre \((-a,-b)\) and radius \(\sqrt{a^2-b^2}\) is
1 \(x^2+y^2+2 a x+2 b y+2 b^2=0\)
2 \(x^2+y^2-2 a x-2 b y-2 b^2=0\)
3 \(x^2+y^2-2 a x-2 b y+2 b^2=0\)
4 \(x^2+y^2-2 a x+2 b y+2 a^2=0\)
Explanation:
A Given that, Center \(=(-a,-b)\) Radius \(=\sqrt{a^2-b^2}\) We know that, general equation of a circle is - \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{r}^2\) Where, \((\mathrm{h}, \mathrm{k})\) centre and \(\mathrm{r}=\) radius \((x+a)^2+(y+b)^2=a^2-b^2\) \(x^2+a^2+2 a x+y^2+b^2+2 y b=a^2-b^2\) \(x^2+y^2+2 a x+2 y b+2 b^2=0 .\)
