NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Conic Section
119627
If the straight line \(3 x+4 y=k\) touches the circle \(x^2+y^2=16 x\), then the values of \(k\) are
1 \(16,-64\)
2 16,64
3 \(-16,-64\)
4 \(-16,64\)
Explanation:
D The straight line \(3 x+4 y-k=0\) touches the circle \(x^2-16 x+y^2=0\) The center of the circle is \((8,0)\) and the radius is 8 . The perpendicular distance between the line and the centre of the circle equals the radius. \(\left|\frac{24-\mathrm{k}}{\sqrt{3^2+4^2}}\right|=8\) \(\left|\frac{24-\mathrm{k}}{5}\right|=8\) \(|24-\mathrm{k}|=40\) \(24-\mathrm{k}= \pm 40\) \(\mathrm{k}=-16,64\)
Karnataka CET-2012
Conic Section
119628
Equation of the circle centered at \((4,3)\) touching the circle \(x^2+y^2=1\) externally, is
1 \(x^2+y^2+8 x+6 y+9=0\)
2 \(x^2+y^2-8 x-6 y+9=0\)
3 \(x^2+y^2-8 x+6 y+9=0\)
4 \(x^2+y^2+8 x-6 y+9=0\)
Explanation:
B Given, equation of circle \(\mathrm{x}^2+\mathrm{y}^2=1\) center at \((0,0)\) And, center of another circle is \((4,3)\), and radius ' \(\mathrm{r}\) ' 1 \(C_1 \ldots, \ldots\) \((0,0)\)\(\ldots\) Then, distance between two centres \(\mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1\) \(\sqrt{(4-0)^2+(3-0)^2}=1+\mathrm{r}\) \(\sqrt{16+9}=1+\mathrm{r}\) \(5=1+\mathrm{r}\) \(\mathrm{r}=4\) Equation of circle \((x-4)^2+(y-3)^2=r^2\) \((x-4)^2+(y-3)^2=4^2\) \(\mathrm{x}^2+16-8 \mathrm{x}+\mathrm{y}^2+9-6 \mathrm{y}=16\) \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-6 \mathrm{y}+9=0\)
Karnataka CET-2010
Conic Section
119629
Find the equation of chord of the circle \(x^2+y^2=8 x\) bisected at the point \((4,3)\)
1 \(y=3\)
2 \(y=1\)
3 \(y=6\)
4 \(y=7\)
Explanation:
A Given,circle equation: \(x^2+y^2=8 x\) \(x^2-8 x+y^2=0\) \((x-4)^2+y^2=4^2\) \(\text { Therefore the center of the circle is }(4,0)\) \(T=S_1\) \(x(4)+y(3)-4(x+4)=16+9-32\) \(3 y-9=0\) \(y=3\)
BITSAT-2010
Conic Section
119632
The locus of the mid-point of a chord of the circle \(x^2+y^2=4\), which subtends a right angle at the origin is
1 \(x+y=2\)
2 \(x^2+y^2=1\)
3 \(x^2+y^2=2\)
4 \(x+y=1\)
Explanation:
C : Given circle \(x^2+y^2=4\) We know that, equation of a circle centered at the origin is given as \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{r}^2\) By comparing given circle is, \(\mathrm{B}=4\) \(\mathrm{r}^2=4\) \(\mathrm{r}=2\) According to figure we get \(\mathrm{BC}\) \(\sin \frac{\pi}{4}=\frac{\mathrm{BC}}{\mathrm{OB}}\) \(\frac{1}{\sqrt{2}}=\frac{\mathrm{BC}}{2}\) \(\mathrm{BC}=\sqrt{2}\) Let the coordinates of \(\mathrm{C}\) be \((\mathrm{x}, \mathrm{y})\) Then the length of \(O C=\sqrt{x^2+y^2}\) From Pythagoras theorem, \(O B^2=\mathrm{OC}^2+\mathrm{BC}^2\) \(2^2=\left(\sqrt{\mathrm{x}^2+\mathrm{y}^2}\right)^2+(\sqrt{2})^2\) \(4=\mathrm{x}^2+\mathrm{y}^2+2\) \(\mathrm{x}^2+\mathrm{y}^2=2\)
119627
If the straight line \(3 x+4 y=k\) touches the circle \(x^2+y^2=16 x\), then the values of \(k\) are
1 \(16,-64\)
2 16,64
3 \(-16,-64\)
4 \(-16,64\)
Explanation:
D The straight line \(3 x+4 y-k=0\) touches the circle \(x^2-16 x+y^2=0\) The center of the circle is \((8,0)\) and the radius is 8 . The perpendicular distance between the line and the centre of the circle equals the radius. \(\left|\frac{24-\mathrm{k}}{\sqrt{3^2+4^2}}\right|=8\) \(\left|\frac{24-\mathrm{k}}{5}\right|=8\) \(|24-\mathrm{k}|=40\) \(24-\mathrm{k}= \pm 40\) \(\mathrm{k}=-16,64\)
Karnataka CET-2012
Conic Section
119628
Equation of the circle centered at \((4,3)\) touching the circle \(x^2+y^2=1\) externally, is
1 \(x^2+y^2+8 x+6 y+9=0\)
2 \(x^2+y^2-8 x-6 y+9=0\)
3 \(x^2+y^2-8 x+6 y+9=0\)
4 \(x^2+y^2+8 x-6 y+9=0\)
Explanation:
B Given, equation of circle \(\mathrm{x}^2+\mathrm{y}^2=1\) center at \((0,0)\) And, center of another circle is \((4,3)\), and radius ' \(\mathrm{r}\) ' 1 \(C_1 \ldots, \ldots\) \((0,0)\)\(\ldots\) Then, distance between two centres \(\mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1\) \(\sqrt{(4-0)^2+(3-0)^2}=1+\mathrm{r}\) \(\sqrt{16+9}=1+\mathrm{r}\) \(5=1+\mathrm{r}\) \(\mathrm{r}=4\) Equation of circle \((x-4)^2+(y-3)^2=r^2\) \((x-4)^2+(y-3)^2=4^2\) \(\mathrm{x}^2+16-8 \mathrm{x}+\mathrm{y}^2+9-6 \mathrm{y}=16\) \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-6 \mathrm{y}+9=0\)
Karnataka CET-2010
Conic Section
119629
Find the equation of chord of the circle \(x^2+y^2=8 x\) bisected at the point \((4,3)\)
1 \(y=3\)
2 \(y=1\)
3 \(y=6\)
4 \(y=7\)
Explanation:
A Given,circle equation: \(x^2+y^2=8 x\) \(x^2-8 x+y^2=0\) \((x-4)^2+y^2=4^2\) \(\text { Therefore the center of the circle is }(4,0)\) \(T=S_1\) \(x(4)+y(3)-4(x+4)=16+9-32\) \(3 y-9=0\) \(y=3\)
BITSAT-2010
Conic Section
119632
The locus of the mid-point of a chord of the circle \(x^2+y^2=4\), which subtends a right angle at the origin is
1 \(x+y=2\)
2 \(x^2+y^2=1\)
3 \(x^2+y^2=2\)
4 \(x+y=1\)
Explanation:
C : Given circle \(x^2+y^2=4\) We know that, equation of a circle centered at the origin is given as \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{r}^2\) By comparing given circle is, \(\mathrm{B}=4\) \(\mathrm{r}^2=4\) \(\mathrm{r}=2\) According to figure we get \(\mathrm{BC}\) \(\sin \frac{\pi}{4}=\frac{\mathrm{BC}}{\mathrm{OB}}\) \(\frac{1}{\sqrt{2}}=\frac{\mathrm{BC}}{2}\) \(\mathrm{BC}=\sqrt{2}\) Let the coordinates of \(\mathrm{C}\) be \((\mathrm{x}, \mathrm{y})\) Then the length of \(O C=\sqrt{x^2+y^2}\) From Pythagoras theorem, \(O B^2=\mathrm{OC}^2+\mathrm{BC}^2\) \(2^2=\left(\sqrt{\mathrm{x}^2+\mathrm{y}^2}\right)^2+(\sqrt{2})^2\) \(4=\mathrm{x}^2+\mathrm{y}^2+2\) \(\mathrm{x}^2+\mathrm{y}^2=2\)
119627
If the straight line \(3 x+4 y=k\) touches the circle \(x^2+y^2=16 x\), then the values of \(k\) are
1 \(16,-64\)
2 16,64
3 \(-16,-64\)
4 \(-16,64\)
Explanation:
D The straight line \(3 x+4 y-k=0\) touches the circle \(x^2-16 x+y^2=0\) The center of the circle is \((8,0)\) and the radius is 8 . The perpendicular distance between the line and the centre of the circle equals the radius. \(\left|\frac{24-\mathrm{k}}{\sqrt{3^2+4^2}}\right|=8\) \(\left|\frac{24-\mathrm{k}}{5}\right|=8\) \(|24-\mathrm{k}|=40\) \(24-\mathrm{k}= \pm 40\) \(\mathrm{k}=-16,64\)
Karnataka CET-2012
Conic Section
119628
Equation of the circle centered at \((4,3)\) touching the circle \(x^2+y^2=1\) externally, is
1 \(x^2+y^2+8 x+6 y+9=0\)
2 \(x^2+y^2-8 x-6 y+9=0\)
3 \(x^2+y^2-8 x+6 y+9=0\)
4 \(x^2+y^2+8 x-6 y+9=0\)
Explanation:
B Given, equation of circle \(\mathrm{x}^2+\mathrm{y}^2=1\) center at \((0,0)\) And, center of another circle is \((4,3)\), and radius ' \(\mathrm{r}\) ' 1 \(C_1 \ldots, \ldots\) \((0,0)\)\(\ldots\) Then, distance between two centres \(\mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1\) \(\sqrt{(4-0)^2+(3-0)^2}=1+\mathrm{r}\) \(\sqrt{16+9}=1+\mathrm{r}\) \(5=1+\mathrm{r}\) \(\mathrm{r}=4\) Equation of circle \((x-4)^2+(y-3)^2=r^2\) \((x-4)^2+(y-3)^2=4^2\) \(\mathrm{x}^2+16-8 \mathrm{x}+\mathrm{y}^2+9-6 \mathrm{y}=16\) \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-6 \mathrm{y}+9=0\)
Karnataka CET-2010
Conic Section
119629
Find the equation of chord of the circle \(x^2+y^2=8 x\) bisected at the point \((4,3)\)
1 \(y=3\)
2 \(y=1\)
3 \(y=6\)
4 \(y=7\)
Explanation:
A Given,circle equation: \(x^2+y^2=8 x\) \(x^2-8 x+y^2=0\) \((x-4)^2+y^2=4^2\) \(\text { Therefore the center of the circle is }(4,0)\) \(T=S_1\) \(x(4)+y(3)-4(x+4)=16+9-32\) \(3 y-9=0\) \(y=3\)
BITSAT-2010
Conic Section
119632
The locus of the mid-point of a chord of the circle \(x^2+y^2=4\), which subtends a right angle at the origin is
1 \(x+y=2\)
2 \(x^2+y^2=1\)
3 \(x^2+y^2=2\)
4 \(x+y=1\)
Explanation:
C : Given circle \(x^2+y^2=4\) We know that, equation of a circle centered at the origin is given as \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{r}^2\) By comparing given circle is, \(\mathrm{B}=4\) \(\mathrm{r}^2=4\) \(\mathrm{r}=2\) According to figure we get \(\mathrm{BC}\) \(\sin \frac{\pi}{4}=\frac{\mathrm{BC}}{\mathrm{OB}}\) \(\frac{1}{\sqrt{2}}=\frac{\mathrm{BC}}{2}\) \(\mathrm{BC}=\sqrt{2}\) Let the coordinates of \(\mathrm{C}\) be \((\mathrm{x}, \mathrm{y})\) Then the length of \(O C=\sqrt{x^2+y^2}\) From Pythagoras theorem, \(O B^2=\mathrm{OC}^2+\mathrm{BC}^2\) \(2^2=\left(\sqrt{\mathrm{x}^2+\mathrm{y}^2}\right)^2+(\sqrt{2})^2\) \(4=\mathrm{x}^2+\mathrm{y}^2+2\) \(\mathrm{x}^2+\mathrm{y}^2=2\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Conic Section
119627
If the straight line \(3 x+4 y=k\) touches the circle \(x^2+y^2=16 x\), then the values of \(k\) are
1 \(16,-64\)
2 16,64
3 \(-16,-64\)
4 \(-16,64\)
Explanation:
D The straight line \(3 x+4 y-k=0\) touches the circle \(x^2-16 x+y^2=0\) The center of the circle is \((8,0)\) and the radius is 8 . The perpendicular distance between the line and the centre of the circle equals the radius. \(\left|\frac{24-\mathrm{k}}{\sqrt{3^2+4^2}}\right|=8\) \(\left|\frac{24-\mathrm{k}}{5}\right|=8\) \(|24-\mathrm{k}|=40\) \(24-\mathrm{k}= \pm 40\) \(\mathrm{k}=-16,64\)
Karnataka CET-2012
Conic Section
119628
Equation of the circle centered at \((4,3)\) touching the circle \(x^2+y^2=1\) externally, is
1 \(x^2+y^2+8 x+6 y+9=0\)
2 \(x^2+y^2-8 x-6 y+9=0\)
3 \(x^2+y^2-8 x+6 y+9=0\)
4 \(x^2+y^2+8 x-6 y+9=0\)
Explanation:
B Given, equation of circle \(\mathrm{x}^2+\mathrm{y}^2=1\) center at \((0,0)\) And, center of another circle is \((4,3)\), and radius ' \(\mathrm{r}\) ' 1 \(C_1 \ldots, \ldots\) \((0,0)\)\(\ldots\) Then, distance between two centres \(\mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1\) \(\sqrt{(4-0)^2+(3-0)^2}=1+\mathrm{r}\) \(\sqrt{16+9}=1+\mathrm{r}\) \(5=1+\mathrm{r}\) \(\mathrm{r}=4\) Equation of circle \((x-4)^2+(y-3)^2=r^2\) \((x-4)^2+(y-3)^2=4^2\) \(\mathrm{x}^2+16-8 \mathrm{x}+\mathrm{y}^2+9-6 \mathrm{y}=16\) \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-6 \mathrm{y}+9=0\)
Karnataka CET-2010
Conic Section
119629
Find the equation of chord of the circle \(x^2+y^2=8 x\) bisected at the point \((4,3)\)
1 \(y=3\)
2 \(y=1\)
3 \(y=6\)
4 \(y=7\)
Explanation:
A Given,circle equation: \(x^2+y^2=8 x\) \(x^2-8 x+y^2=0\) \((x-4)^2+y^2=4^2\) \(\text { Therefore the center of the circle is }(4,0)\) \(T=S_1\) \(x(4)+y(3)-4(x+4)=16+9-32\) \(3 y-9=0\) \(y=3\)
BITSAT-2010
Conic Section
119632
The locus of the mid-point of a chord of the circle \(x^2+y^2=4\), which subtends a right angle at the origin is
1 \(x+y=2\)
2 \(x^2+y^2=1\)
3 \(x^2+y^2=2\)
4 \(x+y=1\)
Explanation:
C : Given circle \(x^2+y^2=4\) We know that, equation of a circle centered at the origin is given as \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{r}^2\) By comparing given circle is, \(\mathrm{B}=4\) \(\mathrm{r}^2=4\) \(\mathrm{r}=2\) According to figure we get \(\mathrm{BC}\) \(\sin \frac{\pi}{4}=\frac{\mathrm{BC}}{\mathrm{OB}}\) \(\frac{1}{\sqrt{2}}=\frac{\mathrm{BC}}{2}\) \(\mathrm{BC}=\sqrt{2}\) Let the coordinates of \(\mathrm{C}\) be \((\mathrm{x}, \mathrm{y})\) Then the length of \(O C=\sqrt{x^2+y^2}\) From Pythagoras theorem, \(O B^2=\mathrm{OC}^2+\mathrm{BC}^2\) \(2^2=\left(\sqrt{\mathrm{x}^2+\mathrm{y}^2}\right)^2+(\sqrt{2})^2\) \(4=\mathrm{x}^2+\mathrm{y}^2+2\) \(\mathrm{x}^2+\mathrm{y}^2=2\)
