119713
If the origin lies on a diameter of the circle \(x^2+\) \(y^2-4 x-2 y-4=0\), then the equation of the circle passing through the end points of that diameter and the point \((1,2)\) is
B Given circle, \(x^2+y^2-4 x-2 y-4=0\) Centre \((2,1)\) Equation of diameter of circle passes through origin. \(\mathrm{x}-2 \mathrm{y}=0\) Equation of circle through the end points of diameter of circle and line \(\mathrm{x}-2 \mathrm{y}=0\) is \(x^2+y^2-4 x-2 y-4+\lambda(x-2 y)=0\) Since, this passes through \((1,2)\) \(\because \quad 1+4-4-4-4+\lambda(-3)=0\) \(\lambda=\frac{-7}{3}\) Required equation of circle is \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-2 \mathrm{y}-4-\frac{7}{3}(\mathrm{x}-2 \mathrm{y})=0\) \(3\left(\mathrm{x}^2+\mathrm{y}^2\right)-19 \mathrm{x}+8 \mathrm{y}-12=0\) \(3 \mathrm{x}^2+3 \mathrm{y}^2-19 \mathrm{x}+8 \mathrm{y}-12=0\)
TS EAMCET-10.09.2020
Conic Section
119714
The locus of the mid points of the chords of the circle \(x^2-2 x+y^2=0\) drawn from a point \((0,0)\) on it is
1 \(x^2+y^2-x=0\)
2 \(2 x^2+y-2=0\)
3 \(y^2+x-1=0\)
4 \(y+x^2+2 x-3=0\)
Explanation:
A : Given, circle \(x^2-2 x+y^2=0\) Centre c \((1,0)\) The locus of the mid-point of the chords of given circle from point \((0,0)\) is a circle. Whose extremities of diameters is centre and given point. Locus of mid point of the chord is - \((\mathrm{x}-1)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-0)=0\) \(\mathrm{x}^2-\mathrm{x}+\mathrm{y}^2=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}=0\)
TS EAMCET-04.08.2021
Conic Section
119715
The equation of a circle of radius 5 units touching another circle \(x^2+y^2-2 x-4 y-20=0\) at \((5,5)\) is
1 \(x^2+y^2+18 x+16 y-220=0\)
2 \(x^2+y^2-x-y-40=0\)
3 \(x^2+y^2+2 x-3 y-45=0\)
4 \(x^2+y^2-18 x-16 y+120=0\)
Explanation:
D Given circle equation, \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-4 \mathrm{y}-20=0\) has its centre \(\mathrm{C}_1=(-\mathrm{g},-\mathrm{f})=(1,2), \mathrm{c}=-20\) Thus radius \((\mathrm{r})=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{(1)^2+(2)^2+20}\) \(=\sqrt{1+4+20}=\sqrt{25}\) \(\mathrm{r}=5\) This circle touches another circle of radius 5 externally at point \(P(5,5)\). Let its center be \(C_2(\alpha, \beta)\) Clearly, \(\mathrm{P}(5,5)\) is the mid point of \(\mathrm{C}_1\) and \(\mathrm{C}_2\) \(\therefore \quad \frac{1+\alpha}{2}=5 \quad, \frac{2+\beta}{2}=5\) On simplifying we get \(\alpha=9 \quad, \beta=8\) Thus the circle is - \((\mathrm{x}-9)^2+(\mathrm{y}-8)^2=5^2\) \(\mathrm{x}^2+\mathrm{y}^2-18 \mathrm{x}-16 \mathrm{y}+120=0\)Or
TS EAMCET-04.05.2019
Conic Section
119716
If the polar of a point \(P\) with respect to a circle of radius \(r\) which touches the coordinate axes and lies in the first quadrant is \(x+2 y=4 r\), then the point \(P\) is
1 \((\mathrm{r}, 2 \mathrm{r})\)
2 \((2 \mathrm{r}, \mathrm{r})\)
3 \((2 \mathrm{r}, 3 \mathrm{r})\)
4 \((-r, 4 \mathrm{r})\)
Explanation:
C Let \(\mathrm{P}\) be \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) Now, equation of circle of radius \(r\) which touches the coordinate axis and lies in the first quadrant is \((x-r)^2+(y-r)^2=r^2\) \(\Rightarrow \quad x^2+y^2-2 x r-2 y r-r^2=0\) Now, polar of \(\mathrm{P}\) with respect to the circle \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{xr}-2 \mathrm{yr}-\mathrm{r}^2=0\) is \(x_1+x_1-\left(x+x_1\right) r-\left(y+y_1\right) r-r^2=0\) \(\Rightarrow \quad\left(x_1-r\right) x+\left(y_1-r\right) y=\left(x_1+y_1-r\right) r\) But it is given that polar of \(P\) with respect to above circle is \(x+2 y=4 r\) \(\therefore \quad \frac{\mathrm{x}_1-\mathrm{r}}{1}=\frac{\mathrm{y}_1-\mathrm{r}}{2}=\frac{\mathrm{x}_1+\mathrm{y}_1-\mathrm{r}}{4}\) \(\Rightarrow \mathrm{x}_1-\mathrm{r}=\frac{\mathrm{x}_1+\mathrm{y}_1-\mathrm{r}}{4} \text { and } \mathrm{x}_1-\mathrm{r}=\frac{\mathrm{y}_1-\mathrm{r}}{2}\) \(\Rightarrow 4 \mathrm{x}_1-4 \mathrm{r}=\mathrm{x}_1+\mathrm{y}_1-\mathrm{r} \text { and } 2 \mathrm{x}_1-2 \mathrm{r}=\mathrm{y}_1-\mathrm{r}\) \(\Rightarrow 3 \mathrm{x}_1-3 \mathrm{r}=\mathrm{y}_1 \text { and } 2 \mathrm{x}_1-\mathrm{r}=\mathrm{y}_1\) \(\Rightarrow \mathrm{x}_1=2 \mathrm{r} \text { and } \mathrm{y}_1=2 \mathrm{x}_1-\mathrm{r}=4 \mathrm{r}-\mathrm{r}=3 \mathrm{r}\) \(\therefore \mathrm{P} \text { is }(2 \mathrm{r}, 3 \mathrm{r})\)
119713
If the origin lies on a diameter of the circle \(x^2+\) \(y^2-4 x-2 y-4=0\), then the equation of the circle passing through the end points of that diameter and the point \((1,2)\) is
B Given circle, \(x^2+y^2-4 x-2 y-4=0\) Centre \((2,1)\) Equation of diameter of circle passes through origin. \(\mathrm{x}-2 \mathrm{y}=0\) Equation of circle through the end points of diameter of circle and line \(\mathrm{x}-2 \mathrm{y}=0\) is \(x^2+y^2-4 x-2 y-4+\lambda(x-2 y)=0\) Since, this passes through \((1,2)\) \(\because \quad 1+4-4-4-4+\lambda(-3)=0\) \(\lambda=\frac{-7}{3}\) Required equation of circle is \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-2 \mathrm{y}-4-\frac{7}{3}(\mathrm{x}-2 \mathrm{y})=0\) \(3\left(\mathrm{x}^2+\mathrm{y}^2\right)-19 \mathrm{x}+8 \mathrm{y}-12=0\) \(3 \mathrm{x}^2+3 \mathrm{y}^2-19 \mathrm{x}+8 \mathrm{y}-12=0\)
TS EAMCET-10.09.2020
Conic Section
119714
The locus of the mid points of the chords of the circle \(x^2-2 x+y^2=0\) drawn from a point \((0,0)\) on it is
1 \(x^2+y^2-x=0\)
2 \(2 x^2+y-2=0\)
3 \(y^2+x-1=0\)
4 \(y+x^2+2 x-3=0\)
Explanation:
A : Given, circle \(x^2-2 x+y^2=0\) Centre c \((1,0)\) The locus of the mid-point of the chords of given circle from point \((0,0)\) is a circle. Whose extremities of diameters is centre and given point. Locus of mid point of the chord is - \((\mathrm{x}-1)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-0)=0\) \(\mathrm{x}^2-\mathrm{x}+\mathrm{y}^2=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}=0\)
TS EAMCET-04.08.2021
Conic Section
119715
The equation of a circle of radius 5 units touching another circle \(x^2+y^2-2 x-4 y-20=0\) at \((5,5)\) is
1 \(x^2+y^2+18 x+16 y-220=0\)
2 \(x^2+y^2-x-y-40=0\)
3 \(x^2+y^2+2 x-3 y-45=0\)
4 \(x^2+y^2-18 x-16 y+120=0\)
Explanation:
D Given circle equation, \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-4 \mathrm{y}-20=0\) has its centre \(\mathrm{C}_1=(-\mathrm{g},-\mathrm{f})=(1,2), \mathrm{c}=-20\) Thus radius \((\mathrm{r})=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{(1)^2+(2)^2+20}\) \(=\sqrt{1+4+20}=\sqrt{25}\) \(\mathrm{r}=5\) This circle touches another circle of radius 5 externally at point \(P(5,5)\). Let its center be \(C_2(\alpha, \beta)\) Clearly, \(\mathrm{P}(5,5)\) is the mid point of \(\mathrm{C}_1\) and \(\mathrm{C}_2\) \(\therefore \quad \frac{1+\alpha}{2}=5 \quad, \frac{2+\beta}{2}=5\) On simplifying we get \(\alpha=9 \quad, \beta=8\) Thus the circle is - \((\mathrm{x}-9)^2+(\mathrm{y}-8)^2=5^2\) \(\mathrm{x}^2+\mathrm{y}^2-18 \mathrm{x}-16 \mathrm{y}+120=0\)Or
TS EAMCET-04.05.2019
Conic Section
119716
If the polar of a point \(P\) with respect to a circle of radius \(r\) which touches the coordinate axes and lies in the first quadrant is \(x+2 y=4 r\), then the point \(P\) is
1 \((\mathrm{r}, 2 \mathrm{r})\)
2 \((2 \mathrm{r}, \mathrm{r})\)
3 \((2 \mathrm{r}, 3 \mathrm{r})\)
4 \((-r, 4 \mathrm{r})\)
Explanation:
C Let \(\mathrm{P}\) be \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) Now, equation of circle of radius \(r\) which touches the coordinate axis and lies in the first quadrant is \((x-r)^2+(y-r)^2=r^2\) \(\Rightarrow \quad x^2+y^2-2 x r-2 y r-r^2=0\) Now, polar of \(\mathrm{P}\) with respect to the circle \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{xr}-2 \mathrm{yr}-\mathrm{r}^2=0\) is \(x_1+x_1-\left(x+x_1\right) r-\left(y+y_1\right) r-r^2=0\) \(\Rightarrow \quad\left(x_1-r\right) x+\left(y_1-r\right) y=\left(x_1+y_1-r\right) r\) But it is given that polar of \(P\) with respect to above circle is \(x+2 y=4 r\) \(\therefore \quad \frac{\mathrm{x}_1-\mathrm{r}}{1}=\frac{\mathrm{y}_1-\mathrm{r}}{2}=\frac{\mathrm{x}_1+\mathrm{y}_1-\mathrm{r}}{4}\) \(\Rightarrow \mathrm{x}_1-\mathrm{r}=\frac{\mathrm{x}_1+\mathrm{y}_1-\mathrm{r}}{4} \text { and } \mathrm{x}_1-\mathrm{r}=\frac{\mathrm{y}_1-\mathrm{r}}{2}\) \(\Rightarrow 4 \mathrm{x}_1-4 \mathrm{r}=\mathrm{x}_1+\mathrm{y}_1-\mathrm{r} \text { and } 2 \mathrm{x}_1-2 \mathrm{r}=\mathrm{y}_1-\mathrm{r}\) \(\Rightarrow 3 \mathrm{x}_1-3 \mathrm{r}=\mathrm{y}_1 \text { and } 2 \mathrm{x}_1-\mathrm{r}=\mathrm{y}_1\) \(\Rightarrow \mathrm{x}_1=2 \mathrm{r} \text { and } \mathrm{y}_1=2 \mathrm{x}_1-\mathrm{r}=4 \mathrm{r}-\mathrm{r}=3 \mathrm{r}\) \(\therefore \mathrm{P} \text { is }(2 \mathrm{r}, 3 \mathrm{r})\)
119713
If the origin lies on a diameter of the circle \(x^2+\) \(y^2-4 x-2 y-4=0\), then the equation of the circle passing through the end points of that diameter and the point \((1,2)\) is
B Given circle, \(x^2+y^2-4 x-2 y-4=0\) Centre \((2,1)\) Equation of diameter of circle passes through origin. \(\mathrm{x}-2 \mathrm{y}=0\) Equation of circle through the end points of diameter of circle and line \(\mathrm{x}-2 \mathrm{y}=0\) is \(x^2+y^2-4 x-2 y-4+\lambda(x-2 y)=0\) Since, this passes through \((1,2)\) \(\because \quad 1+4-4-4-4+\lambda(-3)=0\) \(\lambda=\frac{-7}{3}\) Required equation of circle is \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-2 \mathrm{y}-4-\frac{7}{3}(\mathrm{x}-2 \mathrm{y})=0\) \(3\left(\mathrm{x}^2+\mathrm{y}^2\right)-19 \mathrm{x}+8 \mathrm{y}-12=0\) \(3 \mathrm{x}^2+3 \mathrm{y}^2-19 \mathrm{x}+8 \mathrm{y}-12=0\)
TS EAMCET-10.09.2020
Conic Section
119714
The locus of the mid points of the chords of the circle \(x^2-2 x+y^2=0\) drawn from a point \((0,0)\) on it is
1 \(x^2+y^2-x=0\)
2 \(2 x^2+y-2=0\)
3 \(y^2+x-1=0\)
4 \(y+x^2+2 x-3=0\)
Explanation:
A : Given, circle \(x^2-2 x+y^2=0\) Centre c \((1,0)\) The locus of the mid-point of the chords of given circle from point \((0,0)\) is a circle. Whose extremities of diameters is centre and given point. Locus of mid point of the chord is - \((\mathrm{x}-1)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-0)=0\) \(\mathrm{x}^2-\mathrm{x}+\mathrm{y}^2=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}=0\)
TS EAMCET-04.08.2021
Conic Section
119715
The equation of a circle of radius 5 units touching another circle \(x^2+y^2-2 x-4 y-20=0\) at \((5,5)\) is
1 \(x^2+y^2+18 x+16 y-220=0\)
2 \(x^2+y^2-x-y-40=0\)
3 \(x^2+y^2+2 x-3 y-45=0\)
4 \(x^2+y^2-18 x-16 y+120=0\)
Explanation:
D Given circle equation, \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-4 \mathrm{y}-20=0\) has its centre \(\mathrm{C}_1=(-\mathrm{g},-\mathrm{f})=(1,2), \mathrm{c}=-20\) Thus radius \((\mathrm{r})=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{(1)^2+(2)^2+20}\) \(=\sqrt{1+4+20}=\sqrt{25}\) \(\mathrm{r}=5\) This circle touches another circle of radius 5 externally at point \(P(5,5)\). Let its center be \(C_2(\alpha, \beta)\) Clearly, \(\mathrm{P}(5,5)\) is the mid point of \(\mathrm{C}_1\) and \(\mathrm{C}_2\) \(\therefore \quad \frac{1+\alpha}{2}=5 \quad, \frac{2+\beta}{2}=5\) On simplifying we get \(\alpha=9 \quad, \beta=8\) Thus the circle is - \((\mathrm{x}-9)^2+(\mathrm{y}-8)^2=5^2\) \(\mathrm{x}^2+\mathrm{y}^2-18 \mathrm{x}-16 \mathrm{y}+120=0\)Or
TS EAMCET-04.05.2019
Conic Section
119716
If the polar of a point \(P\) with respect to a circle of radius \(r\) which touches the coordinate axes and lies in the first quadrant is \(x+2 y=4 r\), then the point \(P\) is
1 \((\mathrm{r}, 2 \mathrm{r})\)
2 \((2 \mathrm{r}, \mathrm{r})\)
3 \((2 \mathrm{r}, 3 \mathrm{r})\)
4 \((-r, 4 \mathrm{r})\)
Explanation:
C Let \(\mathrm{P}\) be \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) Now, equation of circle of radius \(r\) which touches the coordinate axis and lies in the first quadrant is \((x-r)^2+(y-r)^2=r^2\) \(\Rightarrow \quad x^2+y^2-2 x r-2 y r-r^2=0\) Now, polar of \(\mathrm{P}\) with respect to the circle \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{xr}-2 \mathrm{yr}-\mathrm{r}^2=0\) is \(x_1+x_1-\left(x+x_1\right) r-\left(y+y_1\right) r-r^2=0\) \(\Rightarrow \quad\left(x_1-r\right) x+\left(y_1-r\right) y=\left(x_1+y_1-r\right) r\) But it is given that polar of \(P\) with respect to above circle is \(x+2 y=4 r\) \(\therefore \quad \frac{\mathrm{x}_1-\mathrm{r}}{1}=\frac{\mathrm{y}_1-\mathrm{r}}{2}=\frac{\mathrm{x}_1+\mathrm{y}_1-\mathrm{r}}{4}\) \(\Rightarrow \mathrm{x}_1-\mathrm{r}=\frac{\mathrm{x}_1+\mathrm{y}_1-\mathrm{r}}{4} \text { and } \mathrm{x}_1-\mathrm{r}=\frac{\mathrm{y}_1-\mathrm{r}}{2}\) \(\Rightarrow 4 \mathrm{x}_1-4 \mathrm{r}=\mathrm{x}_1+\mathrm{y}_1-\mathrm{r} \text { and } 2 \mathrm{x}_1-2 \mathrm{r}=\mathrm{y}_1-\mathrm{r}\) \(\Rightarrow 3 \mathrm{x}_1-3 \mathrm{r}=\mathrm{y}_1 \text { and } 2 \mathrm{x}_1-\mathrm{r}=\mathrm{y}_1\) \(\Rightarrow \mathrm{x}_1=2 \mathrm{r} \text { and } \mathrm{y}_1=2 \mathrm{x}_1-\mathrm{r}=4 \mathrm{r}-\mathrm{r}=3 \mathrm{r}\) \(\therefore \mathrm{P} \text { is }(2 \mathrm{r}, 3 \mathrm{r})\)
119713
If the origin lies on a diameter of the circle \(x^2+\) \(y^2-4 x-2 y-4=0\), then the equation of the circle passing through the end points of that diameter and the point \((1,2)\) is
B Given circle, \(x^2+y^2-4 x-2 y-4=0\) Centre \((2,1)\) Equation of diameter of circle passes through origin. \(\mathrm{x}-2 \mathrm{y}=0\) Equation of circle through the end points of diameter of circle and line \(\mathrm{x}-2 \mathrm{y}=0\) is \(x^2+y^2-4 x-2 y-4+\lambda(x-2 y)=0\) Since, this passes through \((1,2)\) \(\because \quad 1+4-4-4-4+\lambda(-3)=0\) \(\lambda=\frac{-7}{3}\) Required equation of circle is \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-2 \mathrm{y}-4-\frac{7}{3}(\mathrm{x}-2 \mathrm{y})=0\) \(3\left(\mathrm{x}^2+\mathrm{y}^2\right)-19 \mathrm{x}+8 \mathrm{y}-12=0\) \(3 \mathrm{x}^2+3 \mathrm{y}^2-19 \mathrm{x}+8 \mathrm{y}-12=0\)
TS EAMCET-10.09.2020
Conic Section
119714
The locus of the mid points of the chords of the circle \(x^2-2 x+y^2=0\) drawn from a point \((0,0)\) on it is
1 \(x^2+y^2-x=0\)
2 \(2 x^2+y-2=0\)
3 \(y^2+x-1=0\)
4 \(y+x^2+2 x-3=0\)
Explanation:
A : Given, circle \(x^2-2 x+y^2=0\) Centre c \((1,0)\) The locus of the mid-point of the chords of given circle from point \((0,0)\) is a circle. Whose extremities of diameters is centre and given point. Locus of mid point of the chord is - \((\mathrm{x}-1)(\mathrm{x}-0)+(\mathrm{y}-0)(\mathrm{y}-0)=0\) \(\mathrm{x}^2-\mathrm{x}+\mathrm{y}^2=0\) \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}=0\)
TS EAMCET-04.08.2021
Conic Section
119715
The equation of a circle of radius 5 units touching another circle \(x^2+y^2-2 x-4 y-20=0\) at \((5,5)\) is
1 \(x^2+y^2+18 x+16 y-220=0\)
2 \(x^2+y^2-x-y-40=0\)
3 \(x^2+y^2+2 x-3 y-45=0\)
4 \(x^2+y^2-18 x-16 y+120=0\)
Explanation:
D Given circle equation, \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-4 \mathrm{y}-20=0\) has its centre \(\mathrm{C}_1=(-\mathrm{g},-\mathrm{f})=(1,2), \mathrm{c}=-20\) Thus radius \((\mathrm{r})=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}\) \(=\sqrt{(1)^2+(2)^2+20}\) \(=\sqrt{1+4+20}=\sqrt{25}\) \(\mathrm{r}=5\) This circle touches another circle of radius 5 externally at point \(P(5,5)\). Let its center be \(C_2(\alpha, \beta)\) Clearly, \(\mathrm{P}(5,5)\) is the mid point of \(\mathrm{C}_1\) and \(\mathrm{C}_2\) \(\therefore \quad \frac{1+\alpha}{2}=5 \quad, \frac{2+\beta}{2}=5\) On simplifying we get \(\alpha=9 \quad, \beta=8\) Thus the circle is - \((\mathrm{x}-9)^2+(\mathrm{y}-8)^2=5^2\) \(\mathrm{x}^2+\mathrm{y}^2-18 \mathrm{x}-16 \mathrm{y}+120=0\)Or
TS EAMCET-04.05.2019
Conic Section
119716
If the polar of a point \(P\) with respect to a circle of radius \(r\) which touches the coordinate axes and lies in the first quadrant is \(x+2 y=4 r\), then the point \(P\) is
1 \((\mathrm{r}, 2 \mathrm{r})\)
2 \((2 \mathrm{r}, \mathrm{r})\)
3 \((2 \mathrm{r}, 3 \mathrm{r})\)
4 \((-r, 4 \mathrm{r})\)
Explanation:
C Let \(\mathrm{P}\) be \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) Now, equation of circle of radius \(r\) which touches the coordinate axis and lies in the first quadrant is \((x-r)^2+(y-r)^2=r^2\) \(\Rightarrow \quad x^2+y^2-2 x r-2 y r-r^2=0\) Now, polar of \(\mathrm{P}\) with respect to the circle \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{xr}-2 \mathrm{yr}-\mathrm{r}^2=0\) is \(x_1+x_1-\left(x+x_1\right) r-\left(y+y_1\right) r-r^2=0\) \(\Rightarrow \quad\left(x_1-r\right) x+\left(y_1-r\right) y=\left(x_1+y_1-r\right) r\) But it is given that polar of \(P\) with respect to above circle is \(x+2 y=4 r\) \(\therefore \quad \frac{\mathrm{x}_1-\mathrm{r}}{1}=\frac{\mathrm{y}_1-\mathrm{r}}{2}=\frac{\mathrm{x}_1+\mathrm{y}_1-\mathrm{r}}{4}\) \(\Rightarrow \mathrm{x}_1-\mathrm{r}=\frac{\mathrm{x}_1+\mathrm{y}_1-\mathrm{r}}{4} \text { and } \mathrm{x}_1-\mathrm{r}=\frac{\mathrm{y}_1-\mathrm{r}}{2}\) \(\Rightarrow 4 \mathrm{x}_1-4 \mathrm{r}=\mathrm{x}_1+\mathrm{y}_1-\mathrm{r} \text { and } 2 \mathrm{x}_1-2 \mathrm{r}=\mathrm{y}_1-\mathrm{r}\) \(\Rightarrow 3 \mathrm{x}_1-3 \mathrm{r}=\mathrm{y}_1 \text { and } 2 \mathrm{x}_1-\mathrm{r}=\mathrm{y}_1\) \(\Rightarrow \mathrm{x}_1=2 \mathrm{r} \text { and } \mathrm{y}_1=2 \mathrm{x}_1-\mathrm{r}=4 \mathrm{r}-\mathrm{r}=3 \mathrm{r}\) \(\therefore \mathrm{P} \text { is }(2 \mathrm{r}, 3 \mathrm{r})\)
