119782
The point diametrically opposite to the point \(P(1,0)\) on the circle \(x^2+y^2+2 x+4 y-3=0\) is
1 \((3,4)\)
2 \((3,-4)\)
3 \((-3,4)\)
4 \((-3,-4)\) [-2011]
Explanation:
D Given circle equation. \(x^2+y^2+2 x+4 y-3=0\) By comparing above circle equation with general equation of the circle center is \(\left(-\mathrm{g}_1-\mathrm{f}\right)=(-1,-2)\) Let another end of the diameter be (x y) \(\because\) centre is the midpoint of the end points diameter Using midpoint formula \(\frac{x+1}{2}=-1, \frac{y+0}{2}=-2\) \(x=-2-1 y=-4\) \(x=-3\)Hence, point is \((-3,-4)\)
Conic Section
119783
The equation of the circle passing through the points \((1,0)\) and \((0,1)\) and having the smallest radius is
1 \(x^2+y^2+x+y-2=0\)
2 \(x^2+y^2-2 x-2 y+1=0\)
3 \(x^2+y^2-x-y=0\)
4 \(x^2+y^2+2 x+2 y-7=0\) [-2011]
Explanation:
C Given point \((1,0) \&(0,1)\) Let the equation of circle is\(x^2+y^2+2 g x+2 f y+c=0\) At point \((1,0)\) \((1)^2+(0)^2+2 \mathrm{~g} \times 1+0+\mathrm{c}=0\) \(2 \mathrm{~g}+\mathrm{c}+1=0\) At point \((0,1)\) \(0+(1)^2+0+2 \mathrm{f} \times 1+\mathrm{c}=0\) \(2 \mathrm{f}+\mathrm{c}+1=0\) By equation (i) \& (ii) \(g=-\frac{(c+1)}{2} \quad f=\frac{-(c+1)}{2}\) Now, Radius \(\mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{\left(\frac{c+1}{2}\right)^2+\left(\frac{-(c+1)}{2}\right)^2-c}=\sqrt{\frac{(c+1)^2}{2}-c}\) \(r =\sqrt{\frac{c^2+1}{2}} \quad \ldots . \text { (iii) }\) From equation (iii) it is clear \(r\) is minimum when \(c=0\) and minimum value is \(\frac{1}{\sqrt{2}}\) For \(\mathrm{C}=0\) we have \(g=-\frac{1}{2} \quad \text { and } f=\frac{-1}{2}\)On substituting the value of \(g, f\) and \(c\) in general equation of circle we get \(x^2+y^2-x-y=0\)
Conic Section
119784
The length of the diameter of the circle which touches the \(\mathrm{X}\)-axis at the point \((1,0)\) and passes through the point \((2,3)\) is
1 \(\frac{10}{3}\)
2 \(\frac{3}{5}\)
3 \(\frac{6}{5}\)
4 \(\frac{5}{3}\) [-2012]
Explanation:
A Given that the circle passes through points (1, 0 ) and \((2,3)\) thus both coordinates will satisfy the circle equation. \((1-h)^2+k^2=r^2\) Since the circle touches \(\mathrm{x}-\) axis at \((1,0)\) then radius of the circle \(\mathrm{r}=\mathrm{k}\) Here, \((1-\mathrm{h})^2+\mathrm{k}^2=\mathrm{k}^2\) \((1-h)^2=0\) \(\mathrm{~h}=1\) \(\text { and }(2-h)^2+(3-k)^2=k^2\) \(\mathrm{~h}^2+4-4 \mathrm{~h}+9+\mathrm{k}^2-6 \mathrm{k}=\mathrm{k}^2\) \(\mathrm{~h}^2-4 \mathrm{~h}-6 \mathrm{k}+13=0\) \((1)^2-4 \times 1-6 \mathrm{k}+13=0\) \(6 \mathrm{k}=10, \mathrm{k}=\frac{10}{6}=\frac{5}{3}=\frac{5}{3}=\mathrm{r}\) \(\mathrm{D}=2 \times \frac{5}{3}=\frac{10}{3}\) From equation (i) we have \((1)^2-4 \times 1-6 \mathrm{k}+13=0\) \(6 \mathrm{k}=10, \mathrm{k}=\frac{10}{6}=\frac{5}{3}=\frac{5}{3}=\mathrm{r}\)\(\therefore\) Diameter of the circle is \(=2 \mathrm{r}\)
Conic Section
119785
If a circle of radius \(R\) passes through the origin \(O\) and intersects the coordinate axes at \(A\) and \(B\), then the locus of the foot of perpendicular from \(O\) on \(A B\) is
1 \(\left(x^2+y^2\right)^2=4 R^2 x^2 y^2\)
2 \(\left(x^2+y^2\right)^3=4 R^2 x^2 y^2\)
3 \(\left(x^2+y^2\right)(x+y)=R^2 x y\)
4 \(\left(x^2+y^2\right)^2=4 R x^2 y^2\)
Explanation:
B Let us consider the foot of perpendicular be \(\mathrm{P}(\mathrm{h}, \mathrm{k})\). The slope of line \(\mathrm{OP}=\frac{\mathrm{k}}{\mathrm{h}}\) Line \(\mathrm{AB}\) is perpendicular to line \(\mathrm{OP}\). Slope of line \(A B=-\frac{k}{h}\) Now, the equation of line is, \(\mathrm{y}-\mathrm{k}=-\frac{\mathrm{h}}{\mathrm{h}}(\mathrm{x}-\mathrm{h})\) \(-\mathrm{hx}+\mathrm{h}^2=\mathrm{ky}-\mathrm{k}^2\) \(\mathrm{~h}^2+\mathrm{k}^2=\mathrm{ky}+\mathrm{hx}\) \(\frac{\mathrm{hx}+\mathrm{ky}}{\mathrm{h}^2+\mathrm{k}^2}=1\) \(\frac{\mathrm{hx}}{\mathrm{h}^2+\mathrm{k}^2}+\frac{\mathrm{ky}}{\mathrm{h}^2+\mathrm{k}^2}=1\) \(\frac{\mathrm{x}}{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}\right)}+\frac{\mathrm{y}}{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)}=1\) So, point A \(\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}, 0\right)\) And, \(\mathrm{B}\left(0, \frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)\) \(\triangle \mathrm{AOB}\) is a right angled triangle \(\mathrm{So}, \mathrm{AB}\) is one of the diameter of the circle having radius \(\mathrm{R}\) (given). \(2 \mathrm{R}=\mathrm{AB}\) \(\sqrt{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}\right)^2+\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)^2}=2 \mathrm{R}\) \(\left(\mathrm{h}^2+\mathrm{k}^2\right)^3=4 \mathrm{R}^2 \mathrm{~h}^2 \mathrm{k}^2\) On replacing \(\mathrm{h}\) by \(\mathrm{x}\) and \(\mathrm{k}\) by \(\mathrm{y}\) we get, \(\left(x^2+y^2\right)^3=4 R^2 x^2 y^2\) This is required locus.
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Conic Section
119782
The point diametrically opposite to the point \(P(1,0)\) on the circle \(x^2+y^2+2 x+4 y-3=0\) is
1 \((3,4)\)
2 \((3,-4)\)
3 \((-3,4)\)
4 \((-3,-4)\) [-2011]
Explanation:
D Given circle equation. \(x^2+y^2+2 x+4 y-3=0\) By comparing above circle equation with general equation of the circle center is \(\left(-\mathrm{g}_1-\mathrm{f}\right)=(-1,-2)\) Let another end of the diameter be (x y) \(\because\) centre is the midpoint of the end points diameter Using midpoint formula \(\frac{x+1}{2}=-1, \frac{y+0}{2}=-2\) \(x=-2-1 y=-4\) \(x=-3\)Hence, point is \((-3,-4)\)
Conic Section
119783
The equation of the circle passing through the points \((1,0)\) and \((0,1)\) and having the smallest radius is
1 \(x^2+y^2+x+y-2=0\)
2 \(x^2+y^2-2 x-2 y+1=0\)
3 \(x^2+y^2-x-y=0\)
4 \(x^2+y^2+2 x+2 y-7=0\) [-2011]
Explanation:
C Given point \((1,0) \&(0,1)\) Let the equation of circle is\(x^2+y^2+2 g x+2 f y+c=0\) At point \((1,0)\) \((1)^2+(0)^2+2 \mathrm{~g} \times 1+0+\mathrm{c}=0\) \(2 \mathrm{~g}+\mathrm{c}+1=0\) At point \((0,1)\) \(0+(1)^2+0+2 \mathrm{f} \times 1+\mathrm{c}=0\) \(2 \mathrm{f}+\mathrm{c}+1=0\) By equation (i) \& (ii) \(g=-\frac{(c+1)}{2} \quad f=\frac{-(c+1)}{2}\) Now, Radius \(\mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{\left(\frac{c+1}{2}\right)^2+\left(\frac{-(c+1)}{2}\right)^2-c}=\sqrt{\frac{(c+1)^2}{2}-c}\) \(r =\sqrt{\frac{c^2+1}{2}} \quad \ldots . \text { (iii) }\) From equation (iii) it is clear \(r\) is minimum when \(c=0\) and minimum value is \(\frac{1}{\sqrt{2}}\) For \(\mathrm{C}=0\) we have \(g=-\frac{1}{2} \quad \text { and } f=\frac{-1}{2}\)On substituting the value of \(g, f\) and \(c\) in general equation of circle we get \(x^2+y^2-x-y=0\)
Conic Section
119784
The length of the diameter of the circle which touches the \(\mathrm{X}\)-axis at the point \((1,0)\) and passes through the point \((2,3)\) is
1 \(\frac{10}{3}\)
2 \(\frac{3}{5}\)
3 \(\frac{6}{5}\)
4 \(\frac{5}{3}\) [-2012]
Explanation:
A Given that the circle passes through points (1, 0 ) and \((2,3)\) thus both coordinates will satisfy the circle equation. \((1-h)^2+k^2=r^2\) Since the circle touches \(\mathrm{x}-\) axis at \((1,0)\) then radius of the circle \(\mathrm{r}=\mathrm{k}\) Here, \((1-\mathrm{h})^2+\mathrm{k}^2=\mathrm{k}^2\) \((1-h)^2=0\) \(\mathrm{~h}=1\) \(\text { and }(2-h)^2+(3-k)^2=k^2\) \(\mathrm{~h}^2+4-4 \mathrm{~h}+9+\mathrm{k}^2-6 \mathrm{k}=\mathrm{k}^2\) \(\mathrm{~h}^2-4 \mathrm{~h}-6 \mathrm{k}+13=0\) \((1)^2-4 \times 1-6 \mathrm{k}+13=0\) \(6 \mathrm{k}=10, \mathrm{k}=\frac{10}{6}=\frac{5}{3}=\frac{5}{3}=\mathrm{r}\) \(\mathrm{D}=2 \times \frac{5}{3}=\frac{10}{3}\) From equation (i) we have \((1)^2-4 \times 1-6 \mathrm{k}+13=0\) \(6 \mathrm{k}=10, \mathrm{k}=\frac{10}{6}=\frac{5}{3}=\frac{5}{3}=\mathrm{r}\)\(\therefore\) Diameter of the circle is \(=2 \mathrm{r}\)
Conic Section
119785
If a circle of radius \(R\) passes through the origin \(O\) and intersects the coordinate axes at \(A\) and \(B\), then the locus of the foot of perpendicular from \(O\) on \(A B\) is
1 \(\left(x^2+y^2\right)^2=4 R^2 x^2 y^2\)
2 \(\left(x^2+y^2\right)^3=4 R^2 x^2 y^2\)
3 \(\left(x^2+y^2\right)(x+y)=R^2 x y\)
4 \(\left(x^2+y^2\right)^2=4 R x^2 y^2\)
Explanation:
B Let us consider the foot of perpendicular be \(\mathrm{P}(\mathrm{h}, \mathrm{k})\). The slope of line \(\mathrm{OP}=\frac{\mathrm{k}}{\mathrm{h}}\) Line \(\mathrm{AB}\) is perpendicular to line \(\mathrm{OP}\). Slope of line \(A B=-\frac{k}{h}\) Now, the equation of line is, \(\mathrm{y}-\mathrm{k}=-\frac{\mathrm{h}}{\mathrm{h}}(\mathrm{x}-\mathrm{h})\) \(-\mathrm{hx}+\mathrm{h}^2=\mathrm{ky}-\mathrm{k}^2\) \(\mathrm{~h}^2+\mathrm{k}^2=\mathrm{ky}+\mathrm{hx}\) \(\frac{\mathrm{hx}+\mathrm{ky}}{\mathrm{h}^2+\mathrm{k}^2}=1\) \(\frac{\mathrm{hx}}{\mathrm{h}^2+\mathrm{k}^2}+\frac{\mathrm{ky}}{\mathrm{h}^2+\mathrm{k}^2}=1\) \(\frac{\mathrm{x}}{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}\right)}+\frac{\mathrm{y}}{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)}=1\) So, point A \(\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}, 0\right)\) And, \(\mathrm{B}\left(0, \frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)\) \(\triangle \mathrm{AOB}\) is a right angled triangle \(\mathrm{So}, \mathrm{AB}\) is one of the diameter of the circle having radius \(\mathrm{R}\) (given). \(2 \mathrm{R}=\mathrm{AB}\) \(\sqrt{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}\right)^2+\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)^2}=2 \mathrm{R}\) \(\left(\mathrm{h}^2+\mathrm{k}^2\right)^3=4 \mathrm{R}^2 \mathrm{~h}^2 \mathrm{k}^2\) On replacing \(\mathrm{h}\) by \(\mathrm{x}\) and \(\mathrm{k}\) by \(\mathrm{y}\) we get, \(\left(x^2+y^2\right)^3=4 R^2 x^2 y^2\) This is required locus.
119782
The point diametrically opposite to the point \(P(1,0)\) on the circle \(x^2+y^2+2 x+4 y-3=0\) is
1 \((3,4)\)
2 \((3,-4)\)
3 \((-3,4)\)
4 \((-3,-4)\) [-2011]
Explanation:
D Given circle equation. \(x^2+y^2+2 x+4 y-3=0\) By comparing above circle equation with general equation of the circle center is \(\left(-\mathrm{g}_1-\mathrm{f}\right)=(-1,-2)\) Let another end of the diameter be (x y) \(\because\) centre is the midpoint of the end points diameter Using midpoint formula \(\frac{x+1}{2}=-1, \frac{y+0}{2}=-2\) \(x=-2-1 y=-4\) \(x=-3\)Hence, point is \((-3,-4)\)
Conic Section
119783
The equation of the circle passing through the points \((1,0)\) and \((0,1)\) and having the smallest radius is
1 \(x^2+y^2+x+y-2=0\)
2 \(x^2+y^2-2 x-2 y+1=0\)
3 \(x^2+y^2-x-y=0\)
4 \(x^2+y^2+2 x+2 y-7=0\) [-2011]
Explanation:
C Given point \((1,0) \&(0,1)\) Let the equation of circle is\(x^2+y^2+2 g x+2 f y+c=0\) At point \((1,0)\) \((1)^2+(0)^2+2 \mathrm{~g} \times 1+0+\mathrm{c}=0\) \(2 \mathrm{~g}+\mathrm{c}+1=0\) At point \((0,1)\) \(0+(1)^2+0+2 \mathrm{f} \times 1+\mathrm{c}=0\) \(2 \mathrm{f}+\mathrm{c}+1=0\) By equation (i) \& (ii) \(g=-\frac{(c+1)}{2} \quad f=\frac{-(c+1)}{2}\) Now, Radius \(\mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{\left(\frac{c+1}{2}\right)^2+\left(\frac{-(c+1)}{2}\right)^2-c}=\sqrt{\frac{(c+1)^2}{2}-c}\) \(r =\sqrt{\frac{c^2+1}{2}} \quad \ldots . \text { (iii) }\) From equation (iii) it is clear \(r\) is minimum when \(c=0\) and minimum value is \(\frac{1}{\sqrt{2}}\) For \(\mathrm{C}=0\) we have \(g=-\frac{1}{2} \quad \text { and } f=\frac{-1}{2}\)On substituting the value of \(g, f\) and \(c\) in general equation of circle we get \(x^2+y^2-x-y=0\)
Conic Section
119784
The length of the diameter of the circle which touches the \(\mathrm{X}\)-axis at the point \((1,0)\) and passes through the point \((2,3)\) is
1 \(\frac{10}{3}\)
2 \(\frac{3}{5}\)
3 \(\frac{6}{5}\)
4 \(\frac{5}{3}\) [-2012]
Explanation:
A Given that the circle passes through points (1, 0 ) and \((2,3)\) thus both coordinates will satisfy the circle equation. \((1-h)^2+k^2=r^2\) Since the circle touches \(\mathrm{x}-\) axis at \((1,0)\) then radius of the circle \(\mathrm{r}=\mathrm{k}\) Here, \((1-\mathrm{h})^2+\mathrm{k}^2=\mathrm{k}^2\) \((1-h)^2=0\) \(\mathrm{~h}=1\) \(\text { and }(2-h)^2+(3-k)^2=k^2\) \(\mathrm{~h}^2+4-4 \mathrm{~h}+9+\mathrm{k}^2-6 \mathrm{k}=\mathrm{k}^2\) \(\mathrm{~h}^2-4 \mathrm{~h}-6 \mathrm{k}+13=0\) \((1)^2-4 \times 1-6 \mathrm{k}+13=0\) \(6 \mathrm{k}=10, \mathrm{k}=\frac{10}{6}=\frac{5}{3}=\frac{5}{3}=\mathrm{r}\) \(\mathrm{D}=2 \times \frac{5}{3}=\frac{10}{3}\) From equation (i) we have \((1)^2-4 \times 1-6 \mathrm{k}+13=0\) \(6 \mathrm{k}=10, \mathrm{k}=\frac{10}{6}=\frac{5}{3}=\frac{5}{3}=\mathrm{r}\)\(\therefore\) Diameter of the circle is \(=2 \mathrm{r}\)
Conic Section
119785
If a circle of radius \(R\) passes through the origin \(O\) and intersects the coordinate axes at \(A\) and \(B\), then the locus of the foot of perpendicular from \(O\) on \(A B\) is
1 \(\left(x^2+y^2\right)^2=4 R^2 x^2 y^2\)
2 \(\left(x^2+y^2\right)^3=4 R^2 x^2 y^2\)
3 \(\left(x^2+y^2\right)(x+y)=R^2 x y\)
4 \(\left(x^2+y^2\right)^2=4 R x^2 y^2\)
Explanation:
B Let us consider the foot of perpendicular be \(\mathrm{P}(\mathrm{h}, \mathrm{k})\). The slope of line \(\mathrm{OP}=\frac{\mathrm{k}}{\mathrm{h}}\) Line \(\mathrm{AB}\) is perpendicular to line \(\mathrm{OP}\). Slope of line \(A B=-\frac{k}{h}\) Now, the equation of line is, \(\mathrm{y}-\mathrm{k}=-\frac{\mathrm{h}}{\mathrm{h}}(\mathrm{x}-\mathrm{h})\) \(-\mathrm{hx}+\mathrm{h}^2=\mathrm{ky}-\mathrm{k}^2\) \(\mathrm{~h}^2+\mathrm{k}^2=\mathrm{ky}+\mathrm{hx}\) \(\frac{\mathrm{hx}+\mathrm{ky}}{\mathrm{h}^2+\mathrm{k}^2}=1\) \(\frac{\mathrm{hx}}{\mathrm{h}^2+\mathrm{k}^2}+\frac{\mathrm{ky}}{\mathrm{h}^2+\mathrm{k}^2}=1\) \(\frac{\mathrm{x}}{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}\right)}+\frac{\mathrm{y}}{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)}=1\) So, point A \(\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}, 0\right)\) And, \(\mathrm{B}\left(0, \frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)\) \(\triangle \mathrm{AOB}\) is a right angled triangle \(\mathrm{So}, \mathrm{AB}\) is one of the diameter of the circle having radius \(\mathrm{R}\) (given). \(2 \mathrm{R}=\mathrm{AB}\) \(\sqrt{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}\right)^2+\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)^2}=2 \mathrm{R}\) \(\left(\mathrm{h}^2+\mathrm{k}^2\right)^3=4 \mathrm{R}^2 \mathrm{~h}^2 \mathrm{k}^2\) On replacing \(\mathrm{h}\) by \(\mathrm{x}\) and \(\mathrm{k}\) by \(\mathrm{y}\) we get, \(\left(x^2+y^2\right)^3=4 R^2 x^2 y^2\) This is required locus.
119782
The point diametrically opposite to the point \(P(1,0)\) on the circle \(x^2+y^2+2 x+4 y-3=0\) is
1 \((3,4)\)
2 \((3,-4)\)
3 \((-3,4)\)
4 \((-3,-4)\) [-2011]
Explanation:
D Given circle equation. \(x^2+y^2+2 x+4 y-3=0\) By comparing above circle equation with general equation of the circle center is \(\left(-\mathrm{g}_1-\mathrm{f}\right)=(-1,-2)\) Let another end of the diameter be (x y) \(\because\) centre is the midpoint of the end points diameter Using midpoint formula \(\frac{x+1}{2}=-1, \frac{y+0}{2}=-2\) \(x=-2-1 y=-4\) \(x=-3\)Hence, point is \((-3,-4)\)
Conic Section
119783
The equation of the circle passing through the points \((1,0)\) and \((0,1)\) and having the smallest radius is
1 \(x^2+y^2+x+y-2=0\)
2 \(x^2+y^2-2 x-2 y+1=0\)
3 \(x^2+y^2-x-y=0\)
4 \(x^2+y^2+2 x+2 y-7=0\) [-2011]
Explanation:
C Given point \((1,0) \&(0,1)\) Let the equation of circle is\(x^2+y^2+2 g x+2 f y+c=0\) At point \((1,0)\) \((1)^2+(0)^2+2 \mathrm{~g} \times 1+0+\mathrm{c}=0\) \(2 \mathrm{~g}+\mathrm{c}+1=0\) At point \((0,1)\) \(0+(1)^2+0+2 \mathrm{f} \times 1+\mathrm{c}=0\) \(2 \mathrm{f}+\mathrm{c}+1=0\) By equation (i) \& (ii) \(g=-\frac{(c+1)}{2} \quad f=\frac{-(c+1)}{2}\) Now, Radius \(\mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{\left(\frac{c+1}{2}\right)^2+\left(\frac{-(c+1)}{2}\right)^2-c}=\sqrt{\frac{(c+1)^2}{2}-c}\) \(r =\sqrt{\frac{c^2+1}{2}} \quad \ldots . \text { (iii) }\) From equation (iii) it is clear \(r\) is minimum when \(c=0\) and minimum value is \(\frac{1}{\sqrt{2}}\) For \(\mathrm{C}=0\) we have \(g=-\frac{1}{2} \quad \text { and } f=\frac{-1}{2}\)On substituting the value of \(g, f\) and \(c\) in general equation of circle we get \(x^2+y^2-x-y=0\)
Conic Section
119784
The length of the diameter of the circle which touches the \(\mathrm{X}\)-axis at the point \((1,0)\) and passes through the point \((2,3)\) is
1 \(\frac{10}{3}\)
2 \(\frac{3}{5}\)
3 \(\frac{6}{5}\)
4 \(\frac{5}{3}\) [-2012]
Explanation:
A Given that the circle passes through points (1, 0 ) and \((2,3)\) thus both coordinates will satisfy the circle equation. \((1-h)^2+k^2=r^2\) Since the circle touches \(\mathrm{x}-\) axis at \((1,0)\) then radius of the circle \(\mathrm{r}=\mathrm{k}\) Here, \((1-\mathrm{h})^2+\mathrm{k}^2=\mathrm{k}^2\) \((1-h)^2=0\) \(\mathrm{~h}=1\) \(\text { and }(2-h)^2+(3-k)^2=k^2\) \(\mathrm{~h}^2+4-4 \mathrm{~h}+9+\mathrm{k}^2-6 \mathrm{k}=\mathrm{k}^2\) \(\mathrm{~h}^2-4 \mathrm{~h}-6 \mathrm{k}+13=0\) \((1)^2-4 \times 1-6 \mathrm{k}+13=0\) \(6 \mathrm{k}=10, \mathrm{k}=\frac{10}{6}=\frac{5}{3}=\frac{5}{3}=\mathrm{r}\) \(\mathrm{D}=2 \times \frac{5}{3}=\frac{10}{3}\) From equation (i) we have \((1)^2-4 \times 1-6 \mathrm{k}+13=0\) \(6 \mathrm{k}=10, \mathrm{k}=\frac{10}{6}=\frac{5}{3}=\frac{5}{3}=\mathrm{r}\)\(\therefore\) Diameter of the circle is \(=2 \mathrm{r}\)
Conic Section
119785
If a circle of radius \(R\) passes through the origin \(O\) and intersects the coordinate axes at \(A\) and \(B\), then the locus of the foot of perpendicular from \(O\) on \(A B\) is
1 \(\left(x^2+y^2\right)^2=4 R^2 x^2 y^2\)
2 \(\left(x^2+y^2\right)^3=4 R^2 x^2 y^2\)
3 \(\left(x^2+y^2\right)(x+y)=R^2 x y\)
4 \(\left(x^2+y^2\right)^2=4 R x^2 y^2\)
Explanation:
B Let us consider the foot of perpendicular be \(\mathrm{P}(\mathrm{h}, \mathrm{k})\). The slope of line \(\mathrm{OP}=\frac{\mathrm{k}}{\mathrm{h}}\) Line \(\mathrm{AB}\) is perpendicular to line \(\mathrm{OP}\). Slope of line \(A B=-\frac{k}{h}\) Now, the equation of line is, \(\mathrm{y}-\mathrm{k}=-\frac{\mathrm{h}}{\mathrm{h}}(\mathrm{x}-\mathrm{h})\) \(-\mathrm{hx}+\mathrm{h}^2=\mathrm{ky}-\mathrm{k}^2\) \(\mathrm{~h}^2+\mathrm{k}^2=\mathrm{ky}+\mathrm{hx}\) \(\frac{\mathrm{hx}+\mathrm{ky}}{\mathrm{h}^2+\mathrm{k}^2}=1\) \(\frac{\mathrm{hx}}{\mathrm{h}^2+\mathrm{k}^2}+\frac{\mathrm{ky}}{\mathrm{h}^2+\mathrm{k}^2}=1\) \(\frac{\mathrm{x}}{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}\right)}+\frac{\mathrm{y}}{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)}=1\) So, point A \(\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}, 0\right)\) And, \(\mathrm{B}\left(0, \frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)\) \(\triangle \mathrm{AOB}\) is a right angled triangle \(\mathrm{So}, \mathrm{AB}\) is one of the diameter of the circle having radius \(\mathrm{R}\) (given). \(2 \mathrm{R}=\mathrm{AB}\) \(\sqrt{\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{~h}}\right)^2+\left(\frac{\mathrm{h}^2+\mathrm{k}^2}{\mathrm{k}}\right)^2}=2 \mathrm{R}\) \(\left(\mathrm{h}^2+\mathrm{k}^2\right)^3=4 \mathrm{R}^2 \mathrm{~h}^2 \mathrm{k}^2\) On replacing \(\mathrm{h}\) by \(\mathrm{x}\) and \(\mathrm{k}\) by \(\mathrm{y}\) we get, \(\left(x^2+y^2\right)^3=4 R^2 x^2 y^2\) This is required locus.
