88442
Given points \(A(0,0), B(0,4)\) and \(O\) as the origin, find the locus of a point \(P\) such that area of triangle \(\mathrm{POB}\) is 2 time the area of triangle POA.
1 \(x^{2}-3 y^{2}=0\)
2 \(x^{2}+3 y^{2}=0\)
3 \(x^{2}-9 y^{2}=0\)
4 \(x^{2}-4 y^{2}=0\)
Explanation:
(C) : Given the points are:- \(\mathrm{O}(0,0), \quad \mathrm{A}(6,0), \quad \mathrm{B}(0,4)\) And area of \(\triangle \mathrm{POB}=2\) Area of \(\triangle \mathrm{POA}\) Area of triangle \(=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|\) \(\text { let, } \quad \mathrm{P}(\mathrm{h}, \mathrm{k})\) So, area of \((\triangle \mathrm{POB})=\frac{1}{2}|\mathrm{~h}(0-4)+0(4-\mathrm{k})+0(\mathrm{k}-0)|\) \(=\frac{1}{2}|-4 h| . \tag{i}\) \(\operatorname{area} \text { of }(\triangle \mathrm{POA}) =\frac{1}{2}|\mathrm{~h}(0-0)+0(0-\mathrm{k})+6(\mathrm{k}-0)|\) \(=\frac{1}{2}|6 \mathrm{k}| \ldots . .(\mathrm{ii}) \tag{ii}\) According to question \(\frac{1}{2}|-4 \mathrm{~h}|=2\left(\frac{1}{2}|6 \mathrm{k}|\right)\) \(\text { Squaring both sides:- }\) \(\frac{1}{4}\left(16 h^{2}\right)=4\left(\frac{1}{4}\left(36 k^{2}\right)\right)\) \(4 h^{2}=36 k^{2}\) \(h^{2}=9 k^{2}\) \(h^{2}-9 k^{2}=0\) \(\text { or } x^{2}-9 y^{2}=0\)
APEAPCET-2021-23.08.2021
Co-Ordinate system
88443
If a circle of a constant radius 6 passes through origin \(O\) and meets the coordinate axes at \(A\) and \(B\), then find the locus of the centroid of triangle \(O A B\).
1 \(x^{2}+y^{2}=4\)
2 \(x^{2}+y^{2}=36\)
3 \(x^{2}+y^{2}=16\)
4 \(x^{2}+y^{2}=6\)
Explanation:
(C): A (a,0), B(0,b) Hypotenuse \(A B\) is diameter of the circle \(\mathrm{AB}=12\) \(\triangle \mathrm{OAB}\), \(\mathrm{OA}^{2}+\mathrm{OB}^{2}=\mathrm{AB}^{2}\) \(\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=(12)^{2}\) Let \((\alpha, \beta)\) \(\Delta \mathrm{OAB}=\alpha=\frac{\mathrm{a}}{3}, \beta=\frac{\mathrm{b}}{3}\) \(\mathrm{a}=3 \alpha, \mathrm{b}=3 \beta\) \(\Rightarrow(3 \alpha)^{2}+(3 \beta)^{2}=(12)^{2}\) \(\Rightarrow 9 \alpha^{2}+9 \beta^{2}=144 \Rightarrow \alpha^{2}+\beta^{2}=\frac{144}{9}\) \(\Rightarrow \alpha^{2}+\beta^{2}=16 \Rightarrow x^{2}+y^{2}=16\)
Shift-II]
Co-Ordinate system
88444
A variable line passing through \((l, \mathrm{~m})\) intersects the coordinate axes at the points \(A\) and \(B\). If the lines drawn parallel to \(y\)-axis through \(A\) and parallel to \(x\)-axis through \(B\) meet at \(P\), then the locus of \(P\) is
(A) : We have to find that the locus of \(\mathrm{P}=\) ? Let \(\mathrm{OA}=\mathrm{a}\), and \(\mathrm{OB}=\mathrm{b}\) Then, the coordinate of a point \(p\) is \((a, b)\) equation of a variable line \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) Therefore the locus of a point \(\mathrm{P}\) is \(\frac{l}{\mathrm{x}}+\frac{\mathrm{m}}{\mathrm{y}}=1\) \(\mathrm{P}\left(\frac{l}{\mathrm{x}}+\frac{\mathrm{m}}{\mathrm{y}}=1\right)\)
AP EAMCET-2022-06.07.2022
Co-Ordinate system
88445
A line passing through \(P(4,2)\) cuts the coordinate axes at \(A\) and \(B\) respectively. If \(O\) is the origin, then the locus of the centre of the circum-circle of \(\triangle \mathrm{OAB}\) is
1 \(x^{-1}+y^{-1}=2\)
2 \(2 \mathrm{x}^{-1}+\mathrm{y}^{-1}=1\)
3 \(\mathrm{x}^{-1}+2 \mathrm{y}^{-1}=1\)
4 \(2 \mathrm{x}^{-1}+3 \mathrm{y}^{-1}=1\)
Explanation:
(B) : We have to find that lotus of the center of the circum - circle pf \(\triangle \mathrm{OAB}\) Let alien cuts the coordinate axes at A and B respectively is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) So, \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{B}(0, \mathrm{~b})\) and line (i) passes through \(\mathrm{P}(\) 4,2 ) So, \(\frac{4}{a}+\frac{2}{b}=1 \quad\) Now coordinate of center of the circum circle of \(\triangle \mathrm{OAB}\) is mid - point of hypotenuse of right \(\Delta\) \(\mathrm{OAB}\) and its mid point of \(\mathrm{AB}\). So, centre of the circum circle of \(\triangle \mathrm{OAB}\) is \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) Now, from equation (ii) on taking locus of point \(\left(\frac{a}{2}, \frac{b}{2}\right)\) we get \(\frac{2}{\mathrm{x}}+\frac{1}{\mathrm{y}}=1 \Rightarrow 2 \mathrm{x}^{-1}+\mathrm{y}^{-1}=1\)
88442
Given points \(A(0,0), B(0,4)\) and \(O\) as the origin, find the locus of a point \(P\) such that area of triangle \(\mathrm{POB}\) is 2 time the area of triangle POA.
1 \(x^{2}-3 y^{2}=0\)
2 \(x^{2}+3 y^{2}=0\)
3 \(x^{2}-9 y^{2}=0\)
4 \(x^{2}-4 y^{2}=0\)
Explanation:
(C) : Given the points are:- \(\mathrm{O}(0,0), \quad \mathrm{A}(6,0), \quad \mathrm{B}(0,4)\) And area of \(\triangle \mathrm{POB}=2\) Area of \(\triangle \mathrm{POA}\) Area of triangle \(=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|\) \(\text { let, } \quad \mathrm{P}(\mathrm{h}, \mathrm{k})\) So, area of \((\triangle \mathrm{POB})=\frac{1}{2}|\mathrm{~h}(0-4)+0(4-\mathrm{k})+0(\mathrm{k}-0)|\) \(=\frac{1}{2}|-4 h| . \tag{i}\) \(\operatorname{area} \text { of }(\triangle \mathrm{POA}) =\frac{1}{2}|\mathrm{~h}(0-0)+0(0-\mathrm{k})+6(\mathrm{k}-0)|\) \(=\frac{1}{2}|6 \mathrm{k}| \ldots . .(\mathrm{ii}) \tag{ii}\) According to question \(\frac{1}{2}|-4 \mathrm{~h}|=2\left(\frac{1}{2}|6 \mathrm{k}|\right)\) \(\text { Squaring both sides:- }\) \(\frac{1}{4}\left(16 h^{2}\right)=4\left(\frac{1}{4}\left(36 k^{2}\right)\right)\) \(4 h^{2}=36 k^{2}\) \(h^{2}=9 k^{2}\) \(h^{2}-9 k^{2}=0\) \(\text { or } x^{2}-9 y^{2}=0\)
APEAPCET-2021-23.08.2021
Co-Ordinate system
88443
If a circle of a constant radius 6 passes through origin \(O\) and meets the coordinate axes at \(A\) and \(B\), then find the locus of the centroid of triangle \(O A B\).
1 \(x^{2}+y^{2}=4\)
2 \(x^{2}+y^{2}=36\)
3 \(x^{2}+y^{2}=16\)
4 \(x^{2}+y^{2}=6\)
Explanation:
(C): A (a,0), B(0,b) Hypotenuse \(A B\) is diameter of the circle \(\mathrm{AB}=12\) \(\triangle \mathrm{OAB}\), \(\mathrm{OA}^{2}+\mathrm{OB}^{2}=\mathrm{AB}^{2}\) \(\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=(12)^{2}\) Let \((\alpha, \beta)\) \(\Delta \mathrm{OAB}=\alpha=\frac{\mathrm{a}}{3}, \beta=\frac{\mathrm{b}}{3}\) \(\mathrm{a}=3 \alpha, \mathrm{b}=3 \beta\) \(\Rightarrow(3 \alpha)^{2}+(3 \beta)^{2}=(12)^{2}\) \(\Rightarrow 9 \alpha^{2}+9 \beta^{2}=144 \Rightarrow \alpha^{2}+\beta^{2}=\frac{144}{9}\) \(\Rightarrow \alpha^{2}+\beta^{2}=16 \Rightarrow x^{2}+y^{2}=16\)
Shift-II]
Co-Ordinate system
88444
A variable line passing through \((l, \mathrm{~m})\) intersects the coordinate axes at the points \(A\) and \(B\). If the lines drawn parallel to \(y\)-axis through \(A\) and parallel to \(x\)-axis through \(B\) meet at \(P\), then the locus of \(P\) is
(A) : We have to find that the locus of \(\mathrm{P}=\) ? Let \(\mathrm{OA}=\mathrm{a}\), and \(\mathrm{OB}=\mathrm{b}\) Then, the coordinate of a point \(p\) is \((a, b)\) equation of a variable line \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) Therefore the locus of a point \(\mathrm{P}\) is \(\frac{l}{\mathrm{x}}+\frac{\mathrm{m}}{\mathrm{y}}=1\) \(\mathrm{P}\left(\frac{l}{\mathrm{x}}+\frac{\mathrm{m}}{\mathrm{y}}=1\right)\)
AP EAMCET-2022-06.07.2022
Co-Ordinate system
88445
A line passing through \(P(4,2)\) cuts the coordinate axes at \(A\) and \(B\) respectively. If \(O\) is the origin, then the locus of the centre of the circum-circle of \(\triangle \mathrm{OAB}\) is
1 \(x^{-1}+y^{-1}=2\)
2 \(2 \mathrm{x}^{-1}+\mathrm{y}^{-1}=1\)
3 \(\mathrm{x}^{-1}+2 \mathrm{y}^{-1}=1\)
4 \(2 \mathrm{x}^{-1}+3 \mathrm{y}^{-1}=1\)
Explanation:
(B) : We have to find that lotus of the center of the circum - circle pf \(\triangle \mathrm{OAB}\) Let alien cuts the coordinate axes at A and B respectively is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) So, \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{B}(0, \mathrm{~b})\) and line (i) passes through \(\mathrm{P}(\) 4,2 ) So, \(\frac{4}{a}+\frac{2}{b}=1 \quad\) Now coordinate of center of the circum circle of \(\triangle \mathrm{OAB}\) is mid - point of hypotenuse of right \(\Delta\) \(\mathrm{OAB}\) and its mid point of \(\mathrm{AB}\). So, centre of the circum circle of \(\triangle \mathrm{OAB}\) is \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) Now, from equation (ii) on taking locus of point \(\left(\frac{a}{2}, \frac{b}{2}\right)\) we get \(\frac{2}{\mathrm{x}}+\frac{1}{\mathrm{y}}=1 \Rightarrow 2 \mathrm{x}^{-1}+\mathrm{y}^{-1}=1\)
88442
Given points \(A(0,0), B(0,4)\) and \(O\) as the origin, find the locus of a point \(P\) such that area of triangle \(\mathrm{POB}\) is 2 time the area of triangle POA.
1 \(x^{2}-3 y^{2}=0\)
2 \(x^{2}+3 y^{2}=0\)
3 \(x^{2}-9 y^{2}=0\)
4 \(x^{2}-4 y^{2}=0\)
Explanation:
(C) : Given the points are:- \(\mathrm{O}(0,0), \quad \mathrm{A}(6,0), \quad \mathrm{B}(0,4)\) And area of \(\triangle \mathrm{POB}=2\) Area of \(\triangle \mathrm{POA}\) Area of triangle \(=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|\) \(\text { let, } \quad \mathrm{P}(\mathrm{h}, \mathrm{k})\) So, area of \((\triangle \mathrm{POB})=\frac{1}{2}|\mathrm{~h}(0-4)+0(4-\mathrm{k})+0(\mathrm{k}-0)|\) \(=\frac{1}{2}|-4 h| . \tag{i}\) \(\operatorname{area} \text { of }(\triangle \mathrm{POA}) =\frac{1}{2}|\mathrm{~h}(0-0)+0(0-\mathrm{k})+6(\mathrm{k}-0)|\) \(=\frac{1}{2}|6 \mathrm{k}| \ldots . .(\mathrm{ii}) \tag{ii}\) According to question \(\frac{1}{2}|-4 \mathrm{~h}|=2\left(\frac{1}{2}|6 \mathrm{k}|\right)\) \(\text { Squaring both sides:- }\) \(\frac{1}{4}\left(16 h^{2}\right)=4\left(\frac{1}{4}\left(36 k^{2}\right)\right)\) \(4 h^{2}=36 k^{2}\) \(h^{2}=9 k^{2}\) \(h^{2}-9 k^{2}=0\) \(\text { or } x^{2}-9 y^{2}=0\)
APEAPCET-2021-23.08.2021
Co-Ordinate system
88443
If a circle of a constant radius 6 passes through origin \(O\) and meets the coordinate axes at \(A\) and \(B\), then find the locus of the centroid of triangle \(O A B\).
1 \(x^{2}+y^{2}=4\)
2 \(x^{2}+y^{2}=36\)
3 \(x^{2}+y^{2}=16\)
4 \(x^{2}+y^{2}=6\)
Explanation:
(C): A (a,0), B(0,b) Hypotenuse \(A B\) is diameter of the circle \(\mathrm{AB}=12\) \(\triangle \mathrm{OAB}\), \(\mathrm{OA}^{2}+\mathrm{OB}^{2}=\mathrm{AB}^{2}\) \(\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=(12)^{2}\) Let \((\alpha, \beta)\) \(\Delta \mathrm{OAB}=\alpha=\frac{\mathrm{a}}{3}, \beta=\frac{\mathrm{b}}{3}\) \(\mathrm{a}=3 \alpha, \mathrm{b}=3 \beta\) \(\Rightarrow(3 \alpha)^{2}+(3 \beta)^{2}=(12)^{2}\) \(\Rightarrow 9 \alpha^{2}+9 \beta^{2}=144 \Rightarrow \alpha^{2}+\beta^{2}=\frac{144}{9}\) \(\Rightarrow \alpha^{2}+\beta^{2}=16 \Rightarrow x^{2}+y^{2}=16\)
Shift-II]
Co-Ordinate system
88444
A variable line passing through \((l, \mathrm{~m})\) intersects the coordinate axes at the points \(A\) and \(B\). If the lines drawn parallel to \(y\)-axis through \(A\) and parallel to \(x\)-axis through \(B\) meet at \(P\), then the locus of \(P\) is
(A) : We have to find that the locus of \(\mathrm{P}=\) ? Let \(\mathrm{OA}=\mathrm{a}\), and \(\mathrm{OB}=\mathrm{b}\) Then, the coordinate of a point \(p\) is \((a, b)\) equation of a variable line \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) Therefore the locus of a point \(\mathrm{P}\) is \(\frac{l}{\mathrm{x}}+\frac{\mathrm{m}}{\mathrm{y}}=1\) \(\mathrm{P}\left(\frac{l}{\mathrm{x}}+\frac{\mathrm{m}}{\mathrm{y}}=1\right)\)
AP EAMCET-2022-06.07.2022
Co-Ordinate system
88445
A line passing through \(P(4,2)\) cuts the coordinate axes at \(A\) and \(B\) respectively. If \(O\) is the origin, then the locus of the centre of the circum-circle of \(\triangle \mathrm{OAB}\) is
1 \(x^{-1}+y^{-1}=2\)
2 \(2 \mathrm{x}^{-1}+\mathrm{y}^{-1}=1\)
3 \(\mathrm{x}^{-1}+2 \mathrm{y}^{-1}=1\)
4 \(2 \mathrm{x}^{-1}+3 \mathrm{y}^{-1}=1\)
Explanation:
(B) : We have to find that lotus of the center of the circum - circle pf \(\triangle \mathrm{OAB}\) Let alien cuts the coordinate axes at A and B respectively is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) So, \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{B}(0, \mathrm{~b})\) and line (i) passes through \(\mathrm{P}(\) 4,2 ) So, \(\frac{4}{a}+\frac{2}{b}=1 \quad\) Now coordinate of center of the circum circle of \(\triangle \mathrm{OAB}\) is mid - point of hypotenuse of right \(\Delta\) \(\mathrm{OAB}\) and its mid point of \(\mathrm{AB}\). So, centre of the circum circle of \(\triangle \mathrm{OAB}\) is \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) Now, from equation (ii) on taking locus of point \(\left(\frac{a}{2}, \frac{b}{2}\right)\) we get \(\frac{2}{\mathrm{x}}+\frac{1}{\mathrm{y}}=1 \Rightarrow 2 \mathrm{x}^{-1}+\mathrm{y}^{-1}=1\)
88442
Given points \(A(0,0), B(0,4)\) and \(O\) as the origin, find the locus of a point \(P\) such that area of triangle \(\mathrm{POB}\) is 2 time the area of triangle POA.
1 \(x^{2}-3 y^{2}=0\)
2 \(x^{2}+3 y^{2}=0\)
3 \(x^{2}-9 y^{2}=0\)
4 \(x^{2}-4 y^{2}=0\)
Explanation:
(C) : Given the points are:- \(\mathrm{O}(0,0), \quad \mathrm{A}(6,0), \quad \mathrm{B}(0,4)\) And area of \(\triangle \mathrm{POB}=2\) Area of \(\triangle \mathrm{POA}\) Area of triangle \(=\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right|\) \(\text { let, } \quad \mathrm{P}(\mathrm{h}, \mathrm{k})\) So, area of \((\triangle \mathrm{POB})=\frac{1}{2}|\mathrm{~h}(0-4)+0(4-\mathrm{k})+0(\mathrm{k}-0)|\) \(=\frac{1}{2}|-4 h| . \tag{i}\) \(\operatorname{area} \text { of }(\triangle \mathrm{POA}) =\frac{1}{2}|\mathrm{~h}(0-0)+0(0-\mathrm{k})+6(\mathrm{k}-0)|\) \(=\frac{1}{2}|6 \mathrm{k}| \ldots . .(\mathrm{ii}) \tag{ii}\) According to question \(\frac{1}{2}|-4 \mathrm{~h}|=2\left(\frac{1}{2}|6 \mathrm{k}|\right)\) \(\text { Squaring both sides:- }\) \(\frac{1}{4}\left(16 h^{2}\right)=4\left(\frac{1}{4}\left(36 k^{2}\right)\right)\) \(4 h^{2}=36 k^{2}\) \(h^{2}=9 k^{2}\) \(h^{2}-9 k^{2}=0\) \(\text { or } x^{2}-9 y^{2}=0\)
APEAPCET-2021-23.08.2021
Co-Ordinate system
88443
If a circle of a constant radius 6 passes through origin \(O\) and meets the coordinate axes at \(A\) and \(B\), then find the locus of the centroid of triangle \(O A B\).
1 \(x^{2}+y^{2}=4\)
2 \(x^{2}+y^{2}=36\)
3 \(x^{2}+y^{2}=16\)
4 \(x^{2}+y^{2}=6\)
Explanation:
(C): A (a,0), B(0,b) Hypotenuse \(A B\) is diameter of the circle \(\mathrm{AB}=12\) \(\triangle \mathrm{OAB}\), \(\mathrm{OA}^{2}+\mathrm{OB}^{2}=\mathrm{AB}^{2}\) \(\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=(12)^{2}\) Let \((\alpha, \beta)\) \(\Delta \mathrm{OAB}=\alpha=\frac{\mathrm{a}}{3}, \beta=\frac{\mathrm{b}}{3}\) \(\mathrm{a}=3 \alpha, \mathrm{b}=3 \beta\) \(\Rightarrow(3 \alpha)^{2}+(3 \beta)^{2}=(12)^{2}\) \(\Rightarrow 9 \alpha^{2}+9 \beta^{2}=144 \Rightarrow \alpha^{2}+\beta^{2}=\frac{144}{9}\) \(\Rightarrow \alpha^{2}+\beta^{2}=16 \Rightarrow x^{2}+y^{2}=16\)
Shift-II]
Co-Ordinate system
88444
A variable line passing through \((l, \mathrm{~m})\) intersects the coordinate axes at the points \(A\) and \(B\). If the lines drawn parallel to \(y\)-axis through \(A\) and parallel to \(x\)-axis through \(B\) meet at \(P\), then the locus of \(P\) is
(A) : We have to find that the locus of \(\mathrm{P}=\) ? Let \(\mathrm{OA}=\mathrm{a}\), and \(\mathrm{OB}=\mathrm{b}\) Then, the coordinate of a point \(p\) is \((a, b)\) equation of a variable line \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) Therefore the locus of a point \(\mathrm{P}\) is \(\frac{l}{\mathrm{x}}+\frac{\mathrm{m}}{\mathrm{y}}=1\) \(\mathrm{P}\left(\frac{l}{\mathrm{x}}+\frac{\mathrm{m}}{\mathrm{y}}=1\right)\)
AP EAMCET-2022-06.07.2022
Co-Ordinate system
88445
A line passing through \(P(4,2)\) cuts the coordinate axes at \(A\) and \(B\) respectively. If \(O\) is the origin, then the locus of the centre of the circum-circle of \(\triangle \mathrm{OAB}\) is
1 \(x^{-1}+y^{-1}=2\)
2 \(2 \mathrm{x}^{-1}+\mathrm{y}^{-1}=1\)
3 \(\mathrm{x}^{-1}+2 \mathrm{y}^{-1}=1\)
4 \(2 \mathrm{x}^{-1}+3 \mathrm{y}^{-1}=1\)
Explanation:
(B) : We have to find that lotus of the center of the circum - circle pf \(\triangle \mathrm{OAB}\) Let alien cuts the coordinate axes at A and B respectively is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) So, \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{B}(0, \mathrm{~b})\) and line (i) passes through \(\mathrm{P}(\) 4,2 ) So, \(\frac{4}{a}+\frac{2}{b}=1 \quad\) Now coordinate of center of the circum circle of \(\triangle \mathrm{OAB}\) is mid - point of hypotenuse of right \(\Delta\) \(\mathrm{OAB}\) and its mid point of \(\mathrm{AB}\). So, centre of the circum circle of \(\triangle \mathrm{OAB}\) is \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) Now, from equation (ii) on taking locus of point \(\left(\frac{a}{2}, \frac{b}{2}\right)\) we get \(\frac{2}{\mathrm{x}}+\frac{1}{\mathrm{y}}=1 \Rightarrow 2 \mathrm{x}^{-1}+\mathrm{y}^{-1}=1\)
