119648
The other end of the diameter through the point \((-1,1)\) on the circle \(x^2+y^2-6 x+4 y-12=\) 0 is
1 \((-7,5)\)
2 \((-7,-5)\)
3 \((7,-5)\)
4 \((7,5)\)
Explanation:
C Given circle, \(x^2+y^2-6 x+4 y-12=0\) Centre, \((-\mathrm{g},-\mathrm{f})\) \(+2 \mathrm{~g}=-6, \quad+2 \mathrm{f}=4\) \(g=-3, \quad f=+2\) So, centre of the circle \((3,-2)\) According to equation one end of the diameter is \((-1,1)\) Let the other end of the diameter be \((\alpha, \beta)\) \(\therefore \quad \frac{\alpha-1}{2}=3, \frac{\beta+1}{2}=-2\) \(\alpha-1=6, \quad \beta+1=-4\) \(\alpha=7, \quad \beta=-5\) \(\therefore \quad\) Other end of the diameter is \((7,-5)\).
BCECE-2012
Conic Section
119649
Observe the following statements: I. The circle \(x^2+y^2-6 x-4 y-7=0\) touches \(y\) axis II. The circle \(x^2+y^2+6 x+4 y-7=0\) touches \(x\) axis Then, which of the following statements is/are correct?
1 Both I and II
2 Neither I nor II
3 Only I
4 Only II
Explanation:
B \(x^2+y^2-6 x-4 y-7=0\) A general equation of circle is - \((x-h)^2+(y-k)^2=a^2\) If circle touches \(y-\) axis, then \(h=a\) \(\begin{array}{ll} \therefore \quad & (x-h)^2+(y-k)^2=a^2 \\ & (x-3)^2+(y-2)^2=(2 \sqrt{5})^2 \\ & \\ & h=2 \sqrt{5} \\ & h=3 \end{array}\) Hence condition does not satisfy, which means \(x^2+y^2-6 x-4 y-7=0\) does not touch \(y\)-axis \(\begin{aligned} & x^2+y^2+6 x+4 y-7=0 \\ & (x+3)^2+(y+2)^2-(2 \sqrt{5})^2=0 \end{aligned}\) If circle touches \(x\)-axis then, As \(\begin{aligned} & \mathrm{k}=\mathrm{a} \\ & \mathrm{k}=-2 \\ & \mathrm{a}=2 \sqrt{5} \end{aligned}\) Hence, condition does not satisfy which means \(x^2+y^2+6 x+4 y-7=0\) does not touch x -axis.
BCECE-2010
Conic Section
119652
The point on the straight line \(y=2 x+11\) which is nearest to the circle \(16\left(\mathrm{x}^2+\mathrm{y}^2\right)+32 \mathrm{x}-8 \mathrm{y}-50=0\), is
1 \(\left(\frac{9}{2}, 2\right)\)
2 \(\left(\frac{9}{2},-2\right)\)
3 \(\left(-\frac{9}{2}, 2\right)\)
4 \(\left(-\frac{9}{2},-2\right)\)
Explanation:
C Let required point be \((\alpha, \beta)\) on the straight line \(y=2 x+11\), which is nearest to the circle \(16\left(\mathrm{x}^2+\mathrm{y}^2\right)+32 \mathrm{x}-8 \mathrm{y}-50=0\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}-\frac{1}{2} \mathrm{y}-\frac{50}{16}=0\) \(\therefore\) Centre of circle \(=\left(-1, \frac{1}{4}\right)\) and radius, \(\sqrt{1+\frac{1}{16}+\frac{50}{16}}=\frac{\sqrt{67}}{4}\) Now, equation of straight line passing through centre \(\left(-1, \frac{1}{4}\right)\) and \((\alpha, \beta)\) is \(\mathrm{y}-\frac{1}{4}=\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right)(\mathrm{x}+1)\) Now, gradient of this straight line \(=\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right)\) Since, straight line (i) is perpendicular to the line \(y=2 x+11\) \(\therefore\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right) \times 2=-1 \quad\left[\because \mathrm{m}_1 \cdot \mathrm{m}_2=-1\right]\) \(2 \beta-\frac{1}{2}=-\alpha-1\) \(2 \beta+\alpha=-1+\frac{1}{2}=\frac{-1}{2}\) \(4 \beta+2 \alpha=-1 \quad \ldots \text { (ii) }\) \(\because\) Point \((\alpha, \beta)\) lies on straight line \(\therefore \quad y=2 x+11\) \(\beta=2 \alpha+11\) \(\beta-2 \alpha=11\) On solving equation (ii) and (iii), we get \(5 \beta=10\) \(\beta=2\) \(2 \alpha=2-11=-9\) and \(\alpha=\frac{-9}{2}\)\(\therefore\) Required points is \(\left(\frac{-9}{2}, 2\right)\)
CG PET- 2015
Conic Section
119653
The parabola \(y^2=4 x\) and the circle \(x^2+y^2-6 x+1=0\) will
1 intersect at exactly one point
2 touch each other at two distinct points
3 touch each other at exactly one point
4 intersect at two distinct points
Explanation:
B Parabola \(\mathrm{y}^2=4 \mathrm{x}\) and the circle \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+1=0\) ...(i) Put the value of \(y^2\) in equation (ii), \(x^2+4 x-6 x+1=0\) \(x^2-2 x+1=0\) \((x-1)^2=0\) \(x=1\) Putting the value of \(x\) in equation (i) \(y^2=4 \times 1\) \(y= \pm 2\)Therefore, they intersect at two point are \((1,2)\) and \((1,-2)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119648
The other end of the diameter through the point \((-1,1)\) on the circle \(x^2+y^2-6 x+4 y-12=\) 0 is
1 \((-7,5)\)
2 \((-7,-5)\)
3 \((7,-5)\)
4 \((7,5)\)
Explanation:
C Given circle, \(x^2+y^2-6 x+4 y-12=0\) Centre, \((-\mathrm{g},-\mathrm{f})\) \(+2 \mathrm{~g}=-6, \quad+2 \mathrm{f}=4\) \(g=-3, \quad f=+2\) So, centre of the circle \((3,-2)\) According to equation one end of the diameter is \((-1,1)\) Let the other end of the diameter be \((\alpha, \beta)\) \(\therefore \quad \frac{\alpha-1}{2}=3, \frac{\beta+1}{2}=-2\) \(\alpha-1=6, \quad \beta+1=-4\) \(\alpha=7, \quad \beta=-5\) \(\therefore \quad\) Other end of the diameter is \((7,-5)\).
BCECE-2012
Conic Section
119649
Observe the following statements: I. The circle \(x^2+y^2-6 x-4 y-7=0\) touches \(y\) axis II. The circle \(x^2+y^2+6 x+4 y-7=0\) touches \(x\) axis Then, which of the following statements is/are correct?
1 Both I and II
2 Neither I nor II
3 Only I
4 Only II
Explanation:
B \(x^2+y^2-6 x-4 y-7=0\) A general equation of circle is - \((x-h)^2+(y-k)^2=a^2\) If circle touches \(y-\) axis, then \(h=a\) \(\begin{array}{ll} \therefore \quad & (x-h)^2+(y-k)^2=a^2 \\ & (x-3)^2+(y-2)^2=(2 \sqrt{5})^2 \\ & \\ & h=2 \sqrt{5} \\ & h=3 \end{array}\) Hence condition does not satisfy, which means \(x^2+y^2-6 x-4 y-7=0\) does not touch \(y\)-axis \(\begin{aligned} & x^2+y^2+6 x+4 y-7=0 \\ & (x+3)^2+(y+2)^2-(2 \sqrt{5})^2=0 \end{aligned}\) If circle touches \(x\)-axis then, As \(\begin{aligned} & \mathrm{k}=\mathrm{a} \\ & \mathrm{k}=-2 \\ & \mathrm{a}=2 \sqrt{5} \end{aligned}\) Hence, condition does not satisfy which means \(x^2+y^2+6 x+4 y-7=0\) does not touch x -axis.
BCECE-2010
Conic Section
119652
The point on the straight line \(y=2 x+11\) which is nearest to the circle \(16\left(\mathrm{x}^2+\mathrm{y}^2\right)+32 \mathrm{x}-8 \mathrm{y}-50=0\), is
1 \(\left(\frac{9}{2}, 2\right)\)
2 \(\left(\frac{9}{2},-2\right)\)
3 \(\left(-\frac{9}{2}, 2\right)\)
4 \(\left(-\frac{9}{2},-2\right)\)
Explanation:
C Let required point be \((\alpha, \beta)\) on the straight line \(y=2 x+11\), which is nearest to the circle \(16\left(\mathrm{x}^2+\mathrm{y}^2\right)+32 \mathrm{x}-8 \mathrm{y}-50=0\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}-\frac{1}{2} \mathrm{y}-\frac{50}{16}=0\) \(\therefore\) Centre of circle \(=\left(-1, \frac{1}{4}\right)\) and radius, \(\sqrt{1+\frac{1}{16}+\frac{50}{16}}=\frac{\sqrt{67}}{4}\) Now, equation of straight line passing through centre \(\left(-1, \frac{1}{4}\right)\) and \((\alpha, \beta)\) is \(\mathrm{y}-\frac{1}{4}=\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right)(\mathrm{x}+1)\) Now, gradient of this straight line \(=\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right)\) Since, straight line (i) is perpendicular to the line \(y=2 x+11\) \(\therefore\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right) \times 2=-1 \quad\left[\because \mathrm{m}_1 \cdot \mathrm{m}_2=-1\right]\) \(2 \beta-\frac{1}{2}=-\alpha-1\) \(2 \beta+\alpha=-1+\frac{1}{2}=\frac{-1}{2}\) \(4 \beta+2 \alpha=-1 \quad \ldots \text { (ii) }\) \(\because\) Point \((\alpha, \beta)\) lies on straight line \(\therefore \quad y=2 x+11\) \(\beta=2 \alpha+11\) \(\beta-2 \alpha=11\) On solving equation (ii) and (iii), we get \(5 \beta=10\) \(\beta=2\) \(2 \alpha=2-11=-9\) and \(\alpha=\frac{-9}{2}\)\(\therefore\) Required points is \(\left(\frac{-9}{2}, 2\right)\)
CG PET- 2015
Conic Section
119653
The parabola \(y^2=4 x\) and the circle \(x^2+y^2-6 x+1=0\) will
1 intersect at exactly one point
2 touch each other at two distinct points
3 touch each other at exactly one point
4 intersect at two distinct points
Explanation:
B Parabola \(\mathrm{y}^2=4 \mathrm{x}\) and the circle \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+1=0\) ...(i) Put the value of \(y^2\) in equation (ii), \(x^2+4 x-6 x+1=0\) \(x^2-2 x+1=0\) \((x-1)^2=0\) \(x=1\) Putting the value of \(x\) in equation (i) \(y^2=4 \times 1\) \(y= \pm 2\)Therefore, they intersect at two point are \((1,2)\) and \((1,-2)\)
119648
The other end of the diameter through the point \((-1,1)\) on the circle \(x^2+y^2-6 x+4 y-12=\) 0 is
1 \((-7,5)\)
2 \((-7,-5)\)
3 \((7,-5)\)
4 \((7,5)\)
Explanation:
C Given circle, \(x^2+y^2-6 x+4 y-12=0\) Centre, \((-\mathrm{g},-\mathrm{f})\) \(+2 \mathrm{~g}=-6, \quad+2 \mathrm{f}=4\) \(g=-3, \quad f=+2\) So, centre of the circle \((3,-2)\) According to equation one end of the diameter is \((-1,1)\) Let the other end of the diameter be \((\alpha, \beta)\) \(\therefore \quad \frac{\alpha-1}{2}=3, \frac{\beta+1}{2}=-2\) \(\alpha-1=6, \quad \beta+1=-4\) \(\alpha=7, \quad \beta=-5\) \(\therefore \quad\) Other end of the diameter is \((7,-5)\).
BCECE-2012
Conic Section
119649
Observe the following statements: I. The circle \(x^2+y^2-6 x-4 y-7=0\) touches \(y\) axis II. The circle \(x^2+y^2+6 x+4 y-7=0\) touches \(x\) axis Then, which of the following statements is/are correct?
1 Both I and II
2 Neither I nor II
3 Only I
4 Only II
Explanation:
B \(x^2+y^2-6 x-4 y-7=0\) A general equation of circle is - \((x-h)^2+(y-k)^2=a^2\) If circle touches \(y-\) axis, then \(h=a\) \(\begin{array}{ll} \therefore \quad & (x-h)^2+(y-k)^2=a^2 \\ & (x-3)^2+(y-2)^2=(2 \sqrt{5})^2 \\ & \\ & h=2 \sqrt{5} \\ & h=3 \end{array}\) Hence condition does not satisfy, which means \(x^2+y^2-6 x-4 y-7=0\) does not touch \(y\)-axis \(\begin{aligned} & x^2+y^2+6 x+4 y-7=0 \\ & (x+3)^2+(y+2)^2-(2 \sqrt{5})^2=0 \end{aligned}\) If circle touches \(x\)-axis then, As \(\begin{aligned} & \mathrm{k}=\mathrm{a} \\ & \mathrm{k}=-2 \\ & \mathrm{a}=2 \sqrt{5} \end{aligned}\) Hence, condition does not satisfy which means \(x^2+y^2+6 x+4 y-7=0\) does not touch x -axis.
BCECE-2010
Conic Section
119652
The point on the straight line \(y=2 x+11\) which is nearest to the circle \(16\left(\mathrm{x}^2+\mathrm{y}^2\right)+32 \mathrm{x}-8 \mathrm{y}-50=0\), is
1 \(\left(\frac{9}{2}, 2\right)\)
2 \(\left(\frac{9}{2},-2\right)\)
3 \(\left(-\frac{9}{2}, 2\right)\)
4 \(\left(-\frac{9}{2},-2\right)\)
Explanation:
C Let required point be \((\alpha, \beta)\) on the straight line \(y=2 x+11\), which is nearest to the circle \(16\left(\mathrm{x}^2+\mathrm{y}^2\right)+32 \mathrm{x}-8 \mathrm{y}-50=0\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}-\frac{1}{2} \mathrm{y}-\frac{50}{16}=0\) \(\therefore\) Centre of circle \(=\left(-1, \frac{1}{4}\right)\) and radius, \(\sqrt{1+\frac{1}{16}+\frac{50}{16}}=\frac{\sqrt{67}}{4}\) Now, equation of straight line passing through centre \(\left(-1, \frac{1}{4}\right)\) and \((\alpha, \beta)\) is \(\mathrm{y}-\frac{1}{4}=\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right)(\mathrm{x}+1)\) Now, gradient of this straight line \(=\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right)\) Since, straight line (i) is perpendicular to the line \(y=2 x+11\) \(\therefore\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right) \times 2=-1 \quad\left[\because \mathrm{m}_1 \cdot \mathrm{m}_2=-1\right]\) \(2 \beta-\frac{1}{2}=-\alpha-1\) \(2 \beta+\alpha=-1+\frac{1}{2}=\frac{-1}{2}\) \(4 \beta+2 \alpha=-1 \quad \ldots \text { (ii) }\) \(\because\) Point \((\alpha, \beta)\) lies on straight line \(\therefore \quad y=2 x+11\) \(\beta=2 \alpha+11\) \(\beta-2 \alpha=11\) On solving equation (ii) and (iii), we get \(5 \beta=10\) \(\beta=2\) \(2 \alpha=2-11=-9\) and \(\alpha=\frac{-9}{2}\)\(\therefore\) Required points is \(\left(\frac{-9}{2}, 2\right)\)
CG PET- 2015
Conic Section
119653
The parabola \(y^2=4 x\) and the circle \(x^2+y^2-6 x+1=0\) will
1 intersect at exactly one point
2 touch each other at two distinct points
3 touch each other at exactly one point
4 intersect at two distinct points
Explanation:
B Parabola \(\mathrm{y}^2=4 \mathrm{x}\) and the circle \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+1=0\) ...(i) Put the value of \(y^2\) in equation (ii), \(x^2+4 x-6 x+1=0\) \(x^2-2 x+1=0\) \((x-1)^2=0\) \(x=1\) Putting the value of \(x\) in equation (i) \(y^2=4 \times 1\) \(y= \pm 2\)Therefore, they intersect at two point are \((1,2)\) and \((1,-2)\)
119648
The other end of the diameter through the point \((-1,1)\) on the circle \(x^2+y^2-6 x+4 y-12=\) 0 is
1 \((-7,5)\)
2 \((-7,-5)\)
3 \((7,-5)\)
4 \((7,5)\)
Explanation:
C Given circle, \(x^2+y^2-6 x+4 y-12=0\) Centre, \((-\mathrm{g},-\mathrm{f})\) \(+2 \mathrm{~g}=-6, \quad+2 \mathrm{f}=4\) \(g=-3, \quad f=+2\) So, centre of the circle \((3,-2)\) According to equation one end of the diameter is \((-1,1)\) Let the other end of the diameter be \((\alpha, \beta)\) \(\therefore \quad \frac{\alpha-1}{2}=3, \frac{\beta+1}{2}=-2\) \(\alpha-1=6, \quad \beta+1=-4\) \(\alpha=7, \quad \beta=-5\) \(\therefore \quad\) Other end of the diameter is \((7,-5)\).
BCECE-2012
Conic Section
119649
Observe the following statements: I. The circle \(x^2+y^2-6 x-4 y-7=0\) touches \(y\) axis II. The circle \(x^2+y^2+6 x+4 y-7=0\) touches \(x\) axis Then, which of the following statements is/are correct?
1 Both I and II
2 Neither I nor II
3 Only I
4 Only II
Explanation:
B \(x^2+y^2-6 x-4 y-7=0\) A general equation of circle is - \((x-h)^2+(y-k)^2=a^2\) If circle touches \(y-\) axis, then \(h=a\) \(\begin{array}{ll} \therefore \quad & (x-h)^2+(y-k)^2=a^2 \\ & (x-3)^2+(y-2)^2=(2 \sqrt{5})^2 \\ & \\ & h=2 \sqrt{5} \\ & h=3 \end{array}\) Hence condition does not satisfy, which means \(x^2+y^2-6 x-4 y-7=0\) does not touch \(y\)-axis \(\begin{aligned} & x^2+y^2+6 x+4 y-7=0 \\ & (x+3)^2+(y+2)^2-(2 \sqrt{5})^2=0 \end{aligned}\) If circle touches \(x\)-axis then, As \(\begin{aligned} & \mathrm{k}=\mathrm{a} \\ & \mathrm{k}=-2 \\ & \mathrm{a}=2 \sqrt{5} \end{aligned}\) Hence, condition does not satisfy which means \(x^2+y^2+6 x+4 y-7=0\) does not touch x -axis.
BCECE-2010
Conic Section
119652
The point on the straight line \(y=2 x+11\) which is nearest to the circle \(16\left(\mathrm{x}^2+\mathrm{y}^2\right)+32 \mathrm{x}-8 \mathrm{y}-50=0\), is
1 \(\left(\frac{9}{2}, 2\right)\)
2 \(\left(\frac{9}{2},-2\right)\)
3 \(\left(-\frac{9}{2}, 2\right)\)
4 \(\left(-\frac{9}{2},-2\right)\)
Explanation:
C Let required point be \((\alpha, \beta)\) on the straight line \(y=2 x+11\), which is nearest to the circle \(16\left(\mathrm{x}^2+\mathrm{y}^2\right)+32 \mathrm{x}-8 \mathrm{y}-50=0\) \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}-\frac{1}{2} \mathrm{y}-\frac{50}{16}=0\) \(\therefore\) Centre of circle \(=\left(-1, \frac{1}{4}\right)\) and radius, \(\sqrt{1+\frac{1}{16}+\frac{50}{16}}=\frac{\sqrt{67}}{4}\) Now, equation of straight line passing through centre \(\left(-1, \frac{1}{4}\right)\) and \((\alpha, \beta)\) is \(\mathrm{y}-\frac{1}{4}=\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right)(\mathrm{x}+1)\) Now, gradient of this straight line \(=\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right)\) Since, straight line (i) is perpendicular to the line \(y=2 x+11\) \(\therefore\left(\frac{\beta-\frac{1}{4}}{\alpha+1}\right) \times 2=-1 \quad\left[\because \mathrm{m}_1 \cdot \mathrm{m}_2=-1\right]\) \(2 \beta-\frac{1}{2}=-\alpha-1\) \(2 \beta+\alpha=-1+\frac{1}{2}=\frac{-1}{2}\) \(4 \beta+2 \alpha=-1 \quad \ldots \text { (ii) }\) \(\because\) Point \((\alpha, \beta)\) lies on straight line \(\therefore \quad y=2 x+11\) \(\beta=2 \alpha+11\) \(\beta-2 \alpha=11\) On solving equation (ii) and (iii), we get \(5 \beta=10\) \(\beta=2\) \(2 \alpha=2-11=-9\) and \(\alpha=\frac{-9}{2}\)\(\therefore\) Required points is \(\left(\frac{-9}{2}, 2\right)\)
CG PET- 2015
Conic Section
119653
The parabola \(y^2=4 x\) and the circle \(x^2+y^2-6 x+1=0\) will
1 intersect at exactly one point
2 touch each other at two distinct points
3 touch each other at exactly one point
4 intersect at two distinct points
Explanation:
B Parabola \(\mathrm{y}^2=4 \mathrm{x}\) and the circle \(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}+1=0\) ...(i) Put the value of \(y^2\) in equation (ii), \(x^2+4 x-6 x+1=0\) \(x^2-2 x+1=0\) \((x-1)^2=0\) \(x=1\) Putting the value of \(x\) in equation (i) \(y^2=4 \times 1\) \(y= \pm 2\)Therefore, they intersect at two point are \((1,2)\) and \((1,-2)\)