119663
The equation of a circle with centre \((5,4)\) and touching the \(\mathbf{y}\)-axis is
1 \(x^2+y^2-10 x-8 y-16=0\)
2 \(x^2+y^2-10 x-8 y-61=0\)
3 \(x^2+y^2+10 x+8 y+16=0\)
4 \(x^2+y^2-10 x-8 y+16=0\)
Explanation:
D Given, Centre \((5,4)\) circle touches \(\mathrm{Y}\)-axis So, radius \(=5\) units Equation of circle, \((x-5)^2+(y-4)^2=5^2\) \(x^2+y^2-10 x-8 y+16=0\)
AP EAMCET-17.09.2020
Conic Section
119664
A straight line passes through a fixed point \((h, g)\). The locus of the foot of the perpendicular on it drawn from the origin is
1 a straight line
2 an ellipse
3 a parabola
4 a circle
Explanation:
D Given, line passes through point \((\mathrm{h}, \mathrm{g})\) line equation passing through a point :- \(y-y_1=m\left(x-x_1\right)\) \(y-g=m(x-h)\) Now let the slope of line perpendicular to (i) be \(\mathrm{m}_1\). Thus \(\mathrm{m}_1=-\frac{1}{\mathrm{~m}} \rightarrow \text { origin }\) \((0,0) \Rightarrow\left(x_1, y_1\right)\) \((y-0)=\left(-\frac{1}{m}\right)(x-0)\) \(m y=-x\) \(m=-\frac{x}{y}\) put this value in equation (i) :- \(y-g=\left(-\frac{x}{y}\right)(x-h)\) \(y^2-g y=-x^2+x h\) \(x^2+y^2-x h-g y=0\)This equation represents a circle.
SCRA-2015
Conic Section
119665
The radius of the circle given by \(x^2+y^2+z^2+\) \(2 x-2 y-4 z-19=0=x+2 y+2 z+7\), is
1 4
2 3
3 2
4 1
Explanation:
B Given that, radius of circle \(x^2+y^2+z^2+2 x-2 y-4 z-19=0=x+2 y+2 z+7\) Centre of the sphere is \((c)=(-1,1,2)\) Radius of the sphere is \((\mathrm{r})=\sqrt{(7)^2+(-1)^2+(-2)^2+19}=5\) Now distance of the given plane from the centre is \(\mathrm{cp}=\frac{-1+2+4+7}{\sqrt{1+4+4}}=4\) Hence, radius of the circle is \(=\sqrt{\mathrm{r}^2-\mathrm{cp}^2}\) \(=\sqrt{5^2-4^2}\) \(=3\)
AP EAMCET-2011
Conic Section
119666
The radius of the circle whose center lies at (1,2) while cutting the circle \(x^2+y^2+4 x+16 y-\) \(30=0\) orthogonally, is \(\qquad\) units.
1 \(\sqrt{41}\)
2 \(\sqrt{31}\)
3 \(\sqrt{21}\)
4 \(\sqrt{11}\)
Explanation:
D : Given, equation of circle \(x^2+y^2+4 x+16 y-30=0\) \(2 g=4 \quad 2 f=16\) \(g=2 \quad f=8\) \(C_1=(-2,-8)\) \(R_1=\sqrt{(-2)^2+(-8)^2+30}\) \(\quad=\sqrt{4+64+30}=\sqrt{98}\) \(C_2=(1,2)\) \(d=C_1 C_2=\sqrt{(-2-1)^2+(-8-2)^2}\) \(\quad=\sqrt{9+100}=\sqrt{109}\) \(R_1^2+R_2^2=\left(C_1 C_2\right)^2\) \(98+R_2^2=109\) \(R_2^2=109-98\) \(R_2^2=11\) \(R_2=\sqrt{11}\)
119663
The equation of a circle with centre \((5,4)\) and touching the \(\mathbf{y}\)-axis is
1 \(x^2+y^2-10 x-8 y-16=0\)
2 \(x^2+y^2-10 x-8 y-61=0\)
3 \(x^2+y^2+10 x+8 y+16=0\)
4 \(x^2+y^2-10 x-8 y+16=0\)
Explanation:
D Given, Centre \((5,4)\) circle touches \(\mathrm{Y}\)-axis So, radius \(=5\) units Equation of circle, \((x-5)^2+(y-4)^2=5^2\) \(x^2+y^2-10 x-8 y+16=0\)
AP EAMCET-17.09.2020
Conic Section
119664
A straight line passes through a fixed point \((h, g)\). The locus of the foot of the perpendicular on it drawn from the origin is
1 a straight line
2 an ellipse
3 a parabola
4 a circle
Explanation:
D Given, line passes through point \((\mathrm{h}, \mathrm{g})\) line equation passing through a point :- \(y-y_1=m\left(x-x_1\right)\) \(y-g=m(x-h)\) Now let the slope of line perpendicular to (i) be \(\mathrm{m}_1\). Thus \(\mathrm{m}_1=-\frac{1}{\mathrm{~m}} \rightarrow \text { origin }\) \((0,0) \Rightarrow\left(x_1, y_1\right)\) \((y-0)=\left(-\frac{1}{m}\right)(x-0)\) \(m y=-x\) \(m=-\frac{x}{y}\) put this value in equation (i) :- \(y-g=\left(-\frac{x}{y}\right)(x-h)\) \(y^2-g y=-x^2+x h\) \(x^2+y^2-x h-g y=0\)This equation represents a circle.
SCRA-2015
Conic Section
119665
The radius of the circle given by \(x^2+y^2+z^2+\) \(2 x-2 y-4 z-19=0=x+2 y+2 z+7\), is
1 4
2 3
3 2
4 1
Explanation:
B Given that, radius of circle \(x^2+y^2+z^2+2 x-2 y-4 z-19=0=x+2 y+2 z+7\) Centre of the sphere is \((c)=(-1,1,2)\) Radius of the sphere is \((\mathrm{r})=\sqrt{(7)^2+(-1)^2+(-2)^2+19}=5\) Now distance of the given plane from the centre is \(\mathrm{cp}=\frac{-1+2+4+7}{\sqrt{1+4+4}}=4\) Hence, radius of the circle is \(=\sqrt{\mathrm{r}^2-\mathrm{cp}^2}\) \(=\sqrt{5^2-4^2}\) \(=3\)
AP EAMCET-2011
Conic Section
119666
The radius of the circle whose center lies at (1,2) while cutting the circle \(x^2+y^2+4 x+16 y-\) \(30=0\) orthogonally, is \(\qquad\) units.
1 \(\sqrt{41}\)
2 \(\sqrt{31}\)
3 \(\sqrt{21}\)
4 \(\sqrt{11}\)
Explanation:
D : Given, equation of circle \(x^2+y^2+4 x+16 y-30=0\) \(2 g=4 \quad 2 f=16\) \(g=2 \quad f=8\) \(C_1=(-2,-8)\) \(R_1=\sqrt{(-2)^2+(-8)^2+30}\) \(\quad=\sqrt{4+64+30}=\sqrt{98}\) \(C_2=(1,2)\) \(d=C_1 C_2=\sqrt{(-2-1)^2+(-8-2)^2}\) \(\quad=\sqrt{9+100}=\sqrt{109}\) \(R_1^2+R_2^2=\left(C_1 C_2\right)^2\) \(98+R_2^2=109\) \(R_2^2=109-98\) \(R_2^2=11\) \(R_2=\sqrt{11}\)
119663
The equation of a circle with centre \((5,4)\) and touching the \(\mathbf{y}\)-axis is
1 \(x^2+y^2-10 x-8 y-16=0\)
2 \(x^2+y^2-10 x-8 y-61=0\)
3 \(x^2+y^2+10 x+8 y+16=0\)
4 \(x^2+y^2-10 x-8 y+16=0\)
Explanation:
D Given, Centre \((5,4)\) circle touches \(\mathrm{Y}\)-axis So, radius \(=5\) units Equation of circle, \((x-5)^2+(y-4)^2=5^2\) \(x^2+y^2-10 x-8 y+16=0\)
AP EAMCET-17.09.2020
Conic Section
119664
A straight line passes through a fixed point \((h, g)\). The locus of the foot of the perpendicular on it drawn from the origin is
1 a straight line
2 an ellipse
3 a parabola
4 a circle
Explanation:
D Given, line passes through point \((\mathrm{h}, \mathrm{g})\) line equation passing through a point :- \(y-y_1=m\left(x-x_1\right)\) \(y-g=m(x-h)\) Now let the slope of line perpendicular to (i) be \(\mathrm{m}_1\). Thus \(\mathrm{m}_1=-\frac{1}{\mathrm{~m}} \rightarrow \text { origin }\) \((0,0) \Rightarrow\left(x_1, y_1\right)\) \((y-0)=\left(-\frac{1}{m}\right)(x-0)\) \(m y=-x\) \(m=-\frac{x}{y}\) put this value in equation (i) :- \(y-g=\left(-\frac{x}{y}\right)(x-h)\) \(y^2-g y=-x^2+x h\) \(x^2+y^2-x h-g y=0\)This equation represents a circle.
SCRA-2015
Conic Section
119665
The radius of the circle given by \(x^2+y^2+z^2+\) \(2 x-2 y-4 z-19=0=x+2 y+2 z+7\), is
1 4
2 3
3 2
4 1
Explanation:
B Given that, radius of circle \(x^2+y^2+z^2+2 x-2 y-4 z-19=0=x+2 y+2 z+7\) Centre of the sphere is \((c)=(-1,1,2)\) Radius of the sphere is \((\mathrm{r})=\sqrt{(7)^2+(-1)^2+(-2)^2+19}=5\) Now distance of the given plane from the centre is \(\mathrm{cp}=\frac{-1+2+4+7}{\sqrt{1+4+4}}=4\) Hence, radius of the circle is \(=\sqrt{\mathrm{r}^2-\mathrm{cp}^2}\) \(=\sqrt{5^2-4^2}\) \(=3\)
AP EAMCET-2011
Conic Section
119666
The radius of the circle whose center lies at (1,2) while cutting the circle \(x^2+y^2+4 x+16 y-\) \(30=0\) orthogonally, is \(\qquad\) units.
1 \(\sqrt{41}\)
2 \(\sqrt{31}\)
3 \(\sqrt{21}\)
4 \(\sqrt{11}\)
Explanation:
D : Given, equation of circle \(x^2+y^2+4 x+16 y-30=0\) \(2 g=4 \quad 2 f=16\) \(g=2 \quad f=8\) \(C_1=(-2,-8)\) \(R_1=\sqrt{(-2)^2+(-8)^2+30}\) \(\quad=\sqrt{4+64+30}=\sqrt{98}\) \(C_2=(1,2)\) \(d=C_1 C_2=\sqrt{(-2-1)^2+(-8-2)^2}\) \(\quad=\sqrt{9+100}=\sqrt{109}\) \(R_1^2+R_2^2=\left(C_1 C_2\right)^2\) \(98+R_2^2=109\) \(R_2^2=109-98\) \(R_2^2=11\) \(R_2=\sqrt{11}\)
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Conic Section
119663
The equation of a circle with centre \((5,4)\) and touching the \(\mathbf{y}\)-axis is
1 \(x^2+y^2-10 x-8 y-16=0\)
2 \(x^2+y^2-10 x-8 y-61=0\)
3 \(x^2+y^2+10 x+8 y+16=0\)
4 \(x^2+y^2-10 x-8 y+16=0\)
Explanation:
D Given, Centre \((5,4)\) circle touches \(\mathrm{Y}\)-axis So, radius \(=5\) units Equation of circle, \((x-5)^2+(y-4)^2=5^2\) \(x^2+y^2-10 x-8 y+16=0\)
AP EAMCET-17.09.2020
Conic Section
119664
A straight line passes through a fixed point \((h, g)\). The locus of the foot of the perpendicular on it drawn from the origin is
1 a straight line
2 an ellipse
3 a parabola
4 a circle
Explanation:
D Given, line passes through point \((\mathrm{h}, \mathrm{g})\) line equation passing through a point :- \(y-y_1=m\left(x-x_1\right)\) \(y-g=m(x-h)\) Now let the slope of line perpendicular to (i) be \(\mathrm{m}_1\). Thus \(\mathrm{m}_1=-\frac{1}{\mathrm{~m}} \rightarrow \text { origin }\) \((0,0) \Rightarrow\left(x_1, y_1\right)\) \((y-0)=\left(-\frac{1}{m}\right)(x-0)\) \(m y=-x\) \(m=-\frac{x}{y}\) put this value in equation (i) :- \(y-g=\left(-\frac{x}{y}\right)(x-h)\) \(y^2-g y=-x^2+x h\) \(x^2+y^2-x h-g y=0\)This equation represents a circle.
SCRA-2015
Conic Section
119665
The radius of the circle given by \(x^2+y^2+z^2+\) \(2 x-2 y-4 z-19=0=x+2 y+2 z+7\), is
1 4
2 3
3 2
4 1
Explanation:
B Given that, radius of circle \(x^2+y^2+z^2+2 x-2 y-4 z-19=0=x+2 y+2 z+7\) Centre of the sphere is \((c)=(-1,1,2)\) Radius of the sphere is \((\mathrm{r})=\sqrt{(7)^2+(-1)^2+(-2)^2+19}=5\) Now distance of the given plane from the centre is \(\mathrm{cp}=\frac{-1+2+4+7}{\sqrt{1+4+4}}=4\) Hence, radius of the circle is \(=\sqrt{\mathrm{r}^2-\mathrm{cp}^2}\) \(=\sqrt{5^2-4^2}\) \(=3\)
AP EAMCET-2011
Conic Section
119666
The radius of the circle whose center lies at (1,2) while cutting the circle \(x^2+y^2+4 x+16 y-\) \(30=0\) orthogonally, is \(\qquad\) units.
1 \(\sqrt{41}\)
2 \(\sqrt{31}\)
3 \(\sqrt{21}\)
4 \(\sqrt{11}\)
Explanation:
D : Given, equation of circle \(x^2+y^2+4 x+16 y-30=0\) \(2 g=4 \quad 2 f=16\) \(g=2 \quad f=8\) \(C_1=(-2,-8)\) \(R_1=\sqrt{(-2)^2+(-8)^2+30}\) \(\quad=\sqrt{4+64+30}=\sqrt{98}\) \(C_2=(1,2)\) \(d=C_1 C_2=\sqrt{(-2-1)^2+(-8-2)^2}\) \(\quad=\sqrt{9+100}=\sqrt{109}\) \(R_1^2+R_2^2=\left(C_1 C_2\right)^2\) \(98+R_2^2=109\) \(R_2^2=109-98\) \(R_2^2=11\) \(R_2=\sqrt{11}\)
