119879
The point of intersection of the common tangents drawn to the circles \(x^2+y^2-4 x-2 y+1=0\) and \(x^2+y^2-6 x-4 y+4=0\) is
1 \(\left(\frac{5}{2}, \frac{3}{2}\right)\)
2 \(\left(\frac{6}{5}, \frac{1}{5}\right)\)
3 \((0,-1)\)
4 \(\left(\frac{12}{5}, \frac{7}{5}\right)\)
Explanation:
C : Center of the circle \(c_1=(2,1)\) radius \(r_1=2\) Center of the circle \(c_2=(3,2)\) radius \(r_2=3\) \(\therefore\) Point of intersection of common tangents. The point which divides \(\mathrm{C}_1 \mathrm{C}_2\) in the ratio \(-2: 3=\left(\frac{-6+6}{-2+3}, \frac{-4+3}{-2+3}\right)=(0,-1)\)
AP EAMCET-22.04.2019
Conic Section
119880
If the length of the tangent from \((f, g)\) to the circle \(x^2+y^2=6\) be twice the length of the tangent from the same point to the circle \(x^2+y^2+3 x+3 y=0\), then \(f^2+g^2+4 f+4 g+2\) is equal to
1 -1
2 1
3 0
4 -2
Explanation:
C Given, \(\mathrm{P}=(\mathrm{f}, \mathrm{g})\) \(\mathrm{S} ; \quad \mathrm{x}^2+\mathrm{y}^2-6=0\) \(\mathrm{~S}^{\prime} ; \quad \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+3 \mathrm{y}=0\) \(\text { According to the equestion, }\) \(\sqrt{\mathrm{S}_{11}}=2 \sqrt{\mathrm{S}_{11}^{\prime}}\) \(\mathrm{S}_{11}=4 \mathrm{~S}_{11}^{\prime}\) \(\left(\mathrm{g}^2+\mathrm{f}^2-6\right)=4\left(\mathrm{~g}^2+\mathrm{f}^2+3 \mathrm{~g}+3 \mathrm{f}\right)\) \(3 \mathrm{~g}^2+3 \mathrm{f}^2+12 \mathrm{~g}+12 \mathrm{f}+6=0\) \(\text { Dividing by } 3 \text { on both sides, }\) \(\mathrm{g}^2+\mathrm{f}^2+4 \mathrm{~g}+4 \mathrm{f}+2=0\)
AP EAMCET-22.09.2020
Conic Section
119882
If the lengths of the tangents drawn from \(P\) to the circles \(x^2+y^2-2 x+4 y-20=0\) and \(x^2+y^2-2 x-8 y+1=0\) are in the ratio 2:1, then the locus of \(P\) is
D We know that, Length of tangent drawn from \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to the circle \(x^2+y^2+2 g x+2 f y+c=0\) is \(\sqrt{x_1^2+y_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\) Let, \(\quad \mathrm{P}(\mathrm{h}, \mathrm{k})\), \(\therefore\) According to the question \(\frac{\sqrt{\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20}}{\sqrt{\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}-8 \mathrm{k}+1}}=\frac{2}{1}\) \(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20=4\left(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}-8 \mathrm{k}+1\right)\) \(\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20=4 \mathrm{~h}^2+4 \mathrm{k}^2-8 \mathrm{~h}-32 \mathrm{k}+4\) \(3 \mathrm{~h}^2+3 \mathrm{k}^2-6 \mathrm{~h}-36 \mathrm{k}+24=0\) \(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}-12 \mathrm{k}+8=0\) \(\therefore \text { Locus of the point p is } \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-12 \mathrm{y}+8=0\)
AP EAMCET-22.04.2018
Conic Section
119883
Two tangents to the circle \(x^2+y^2=4\) at the points \(A\) and \(B\) meet at \(M(-4,0)\). The area of the quadrilateral MAOB, where \(O\) is the origin is
1 \(4 \sqrt{3}\) sq. units
2 \(2 \sqrt{3}\) sq. units
3 \(\sqrt{3}\) sq. units
4 \(3 \sqrt{3}\) sq. units
Explanation:
A \(\mathrm{x}^2+\mathrm{y}^2=4\) \(\mathrm{x}^2+\mathrm{y}^2=(2)^2\) \(\triangle \mathrm{OAM} \& \triangle \mathrm{OBM}\) \(\mathrm{OM}^2=\mathrm{AM}^2+\mathrm{OA}^2\) \((4)^2=(\mathrm{AM})^2+(2)^2\) \(\mathrm{AM}^2=16-4=12\) \(\mathrm{AM}=\sqrt{12}\) \(\text { Similarly } B M=\sqrt{12}\) \(\text { Area of OAMB }=2 \times \text { Area of } \triangle \mathrm{OAM}\) \(=2 \times \frac{1}{2} \times \mathrm{AM} \times \mathrm{OA}\) \(=\sqrt{12} \times 2=\sqrt{2 \times 2 \times 3} \times 2\) \(=2 \times 2 \sqrt{3}=4 \sqrt{3} \text { square unit }\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119879
The point of intersection of the common tangents drawn to the circles \(x^2+y^2-4 x-2 y+1=0\) and \(x^2+y^2-6 x-4 y+4=0\) is
1 \(\left(\frac{5}{2}, \frac{3}{2}\right)\)
2 \(\left(\frac{6}{5}, \frac{1}{5}\right)\)
3 \((0,-1)\)
4 \(\left(\frac{12}{5}, \frac{7}{5}\right)\)
Explanation:
C : Center of the circle \(c_1=(2,1)\) radius \(r_1=2\) Center of the circle \(c_2=(3,2)\) radius \(r_2=3\) \(\therefore\) Point of intersection of common tangents. The point which divides \(\mathrm{C}_1 \mathrm{C}_2\) in the ratio \(-2: 3=\left(\frac{-6+6}{-2+3}, \frac{-4+3}{-2+3}\right)=(0,-1)\)
AP EAMCET-22.04.2019
Conic Section
119880
If the length of the tangent from \((f, g)\) to the circle \(x^2+y^2=6\) be twice the length of the tangent from the same point to the circle \(x^2+y^2+3 x+3 y=0\), then \(f^2+g^2+4 f+4 g+2\) is equal to
1 -1
2 1
3 0
4 -2
Explanation:
C Given, \(\mathrm{P}=(\mathrm{f}, \mathrm{g})\) \(\mathrm{S} ; \quad \mathrm{x}^2+\mathrm{y}^2-6=0\) \(\mathrm{~S}^{\prime} ; \quad \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+3 \mathrm{y}=0\) \(\text { According to the equestion, }\) \(\sqrt{\mathrm{S}_{11}}=2 \sqrt{\mathrm{S}_{11}^{\prime}}\) \(\mathrm{S}_{11}=4 \mathrm{~S}_{11}^{\prime}\) \(\left(\mathrm{g}^2+\mathrm{f}^2-6\right)=4\left(\mathrm{~g}^2+\mathrm{f}^2+3 \mathrm{~g}+3 \mathrm{f}\right)\) \(3 \mathrm{~g}^2+3 \mathrm{f}^2+12 \mathrm{~g}+12 \mathrm{f}+6=0\) \(\text { Dividing by } 3 \text { on both sides, }\) \(\mathrm{g}^2+\mathrm{f}^2+4 \mathrm{~g}+4 \mathrm{f}+2=0\)
AP EAMCET-22.09.2020
Conic Section
119882
If the lengths of the tangents drawn from \(P\) to the circles \(x^2+y^2-2 x+4 y-20=0\) and \(x^2+y^2-2 x-8 y+1=0\) are in the ratio 2:1, then the locus of \(P\) is
D We know that, Length of tangent drawn from \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to the circle \(x^2+y^2+2 g x+2 f y+c=0\) is \(\sqrt{x_1^2+y_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\) Let, \(\quad \mathrm{P}(\mathrm{h}, \mathrm{k})\), \(\therefore\) According to the question \(\frac{\sqrt{\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20}}{\sqrt{\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}-8 \mathrm{k}+1}}=\frac{2}{1}\) \(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20=4\left(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}-8 \mathrm{k}+1\right)\) \(\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20=4 \mathrm{~h}^2+4 \mathrm{k}^2-8 \mathrm{~h}-32 \mathrm{k}+4\) \(3 \mathrm{~h}^2+3 \mathrm{k}^2-6 \mathrm{~h}-36 \mathrm{k}+24=0\) \(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}-12 \mathrm{k}+8=0\) \(\therefore \text { Locus of the point p is } \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-12 \mathrm{y}+8=0\)
AP EAMCET-22.04.2018
Conic Section
119883
Two tangents to the circle \(x^2+y^2=4\) at the points \(A\) and \(B\) meet at \(M(-4,0)\). The area of the quadrilateral MAOB, where \(O\) is the origin is
1 \(4 \sqrt{3}\) sq. units
2 \(2 \sqrt{3}\) sq. units
3 \(\sqrt{3}\) sq. units
4 \(3 \sqrt{3}\) sq. units
Explanation:
A \(\mathrm{x}^2+\mathrm{y}^2=4\) \(\mathrm{x}^2+\mathrm{y}^2=(2)^2\) \(\triangle \mathrm{OAM} \& \triangle \mathrm{OBM}\) \(\mathrm{OM}^2=\mathrm{AM}^2+\mathrm{OA}^2\) \((4)^2=(\mathrm{AM})^2+(2)^2\) \(\mathrm{AM}^2=16-4=12\) \(\mathrm{AM}=\sqrt{12}\) \(\text { Similarly } B M=\sqrt{12}\) \(\text { Area of OAMB }=2 \times \text { Area of } \triangle \mathrm{OAM}\) \(=2 \times \frac{1}{2} \times \mathrm{AM} \times \mathrm{OA}\) \(=\sqrt{12} \times 2=\sqrt{2 \times 2 \times 3} \times 2\) \(=2 \times 2 \sqrt{3}=4 \sqrt{3} \text { square unit }\)
119879
The point of intersection of the common tangents drawn to the circles \(x^2+y^2-4 x-2 y+1=0\) and \(x^2+y^2-6 x-4 y+4=0\) is
1 \(\left(\frac{5}{2}, \frac{3}{2}\right)\)
2 \(\left(\frac{6}{5}, \frac{1}{5}\right)\)
3 \((0,-1)\)
4 \(\left(\frac{12}{5}, \frac{7}{5}\right)\)
Explanation:
C : Center of the circle \(c_1=(2,1)\) radius \(r_1=2\) Center of the circle \(c_2=(3,2)\) radius \(r_2=3\) \(\therefore\) Point of intersection of common tangents. The point which divides \(\mathrm{C}_1 \mathrm{C}_2\) in the ratio \(-2: 3=\left(\frac{-6+6}{-2+3}, \frac{-4+3}{-2+3}\right)=(0,-1)\)
AP EAMCET-22.04.2019
Conic Section
119880
If the length of the tangent from \((f, g)\) to the circle \(x^2+y^2=6\) be twice the length of the tangent from the same point to the circle \(x^2+y^2+3 x+3 y=0\), then \(f^2+g^2+4 f+4 g+2\) is equal to
1 -1
2 1
3 0
4 -2
Explanation:
C Given, \(\mathrm{P}=(\mathrm{f}, \mathrm{g})\) \(\mathrm{S} ; \quad \mathrm{x}^2+\mathrm{y}^2-6=0\) \(\mathrm{~S}^{\prime} ; \quad \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+3 \mathrm{y}=0\) \(\text { According to the equestion, }\) \(\sqrt{\mathrm{S}_{11}}=2 \sqrt{\mathrm{S}_{11}^{\prime}}\) \(\mathrm{S}_{11}=4 \mathrm{~S}_{11}^{\prime}\) \(\left(\mathrm{g}^2+\mathrm{f}^2-6\right)=4\left(\mathrm{~g}^2+\mathrm{f}^2+3 \mathrm{~g}+3 \mathrm{f}\right)\) \(3 \mathrm{~g}^2+3 \mathrm{f}^2+12 \mathrm{~g}+12 \mathrm{f}+6=0\) \(\text { Dividing by } 3 \text { on both sides, }\) \(\mathrm{g}^2+\mathrm{f}^2+4 \mathrm{~g}+4 \mathrm{f}+2=0\)
AP EAMCET-22.09.2020
Conic Section
119882
If the lengths of the tangents drawn from \(P\) to the circles \(x^2+y^2-2 x+4 y-20=0\) and \(x^2+y^2-2 x-8 y+1=0\) are in the ratio 2:1, then the locus of \(P\) is
D We know that, Length of tangent drawn from \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to the circle \(x^2+y^2+2 g x+2 f y+c=0\) is \(\sqrt{x_1^2+y_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\) Let, \(\quad \mathrm{P}(\mathrm{h}, \mathrm{k})\), \(\therefore\) According to the question \(\frac{\sqrt{\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20}}{\sqrt{\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}-8 \mathrm{k}+1}}=\frac{2}{1}\) \(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20=4\left(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}-8 \mathrm{k}+1\right)\) \(\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20=4 \mathrm{~h}^2+4 \mathrm{k}^2-8 \mathrm{~h}-32 \mathrm{k}+4\) \(3 \mathrm{~h}^2+3 \mathrm{k}^2-6 \mathrm{~h}-36 \mathrm{k}+24=0\) \(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}-12 \mathrm{k}+8=0\) \(\therefore \text { Locus of the point p is } \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-12 \mathrm{y}+8=0\)
AP EAMCET-22.04.2018
Conic Section
119883
Two tangents to the circle \(x^2+y^2=4\) at the points \(A\) and \(B\) meet at \(M(-4,0)\). The area of the quadrilateral MAOB, where \(O\) is the origin is
1 \(4 \sqrt{3}\) sq. units
2 \(2 \sqrt{3}\) sq. units
3 \(\sqrt{3}\) sq. units
4 \(3 \sqrt{3}\) sq. units
Explanation:
A \(\mathrm{x}^2+\mathrm{y}^2=4\) \(\mathrm{x}^2+\mathrm{y}^2=(2)^2\) \(\triangle \mathrm{OAM} \& \triangle \mathrm{OBM}\) \(\mathrm{OM}^2=\mathrm{AM}^2+\mathrm{OA}^2\) \((4)^2=(\mathrm{AM})^2+(2)^2\) \(\mathrm{AM}^2=16-4=12\) \(\mathrm{AM}=\sqrt{12}\) \(\text { Similarly } B M=\sqrt{12}\) \(\text { Area of OAMB }=2 \times \text { Area of } \triangle \mathrm{OAM}\) \(=2 \times \frac{1}{2} \times \mathrm{AM} \times \mathrm{OA}\) \(=\sqrt{12} \times 2=\sqrt{2 \times 2 \times 3} \times 2\) \(=2 \times 2 \sqrt{3}=4 \sqrt{3} \text { square unit }\)
119879
The point of intersection of the common tangents drawn to the circles \(x^2+y^2-4 x-2 y+1=0\) and \(x^2+y^2-6 x-4 y+4=0\) is
1 \(\left(\frac{5}{2}, \frac{3}{2}\right)\)
2 \(\left(\frac{6}{5}, \frac{1}{5}\right)\)
3 \((0,-1)\)
4 \(\left(\frac{12}{5}, \frac{7}{5}\right)\)
Explanation:
C : Center of the circle \(c_1=(2,1)\) radius \(r_1=2\) Center of the circle \(c_2=(3,2)\) radius \(r_2=3\) \(\therefore\) Point of intersection of common tangents. The point which divides \(\mathrm{C}_1 \mathrm{C}_2\) in the ratio \(-2: 3=\left(\frac{-6+6}{-2+3}, \frac{-4+3}{-2+3}\right)=(0,-1)\)
AP EAMCET-22.04.2019
Conic Section
119880
If the length of the tangent from \((f, g)\) to the circle \(x^2+y^2=6\) be twice the length of the tangent from the same point to the circle \(x^2+y^2+3 x+3 y=0\), then \(f^2+g^2+4 f+4 g+2\) is equal to
1 -1
2 1
3 0
4 -2
Explanation:
C Given, \(\mathrm{P}=(\mathrm{f}, \mathrm{g})\) \(\mathrm{S} ; \quad \mathrm{x}^2+\mathrm{y}^2-6=0\) \(\mathrm{~S}^{\prime} ; \quad \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+3 \mathrm{y}=0\) \(\text { According to the equestion, }\) \(\sqrt{\mathrm{S}_{11}}=2 \sqrt{\mathrm{S}_{11}^{\prime}}\) \(\mathrm{S}_{11}=4 \mathrm{~S}_{11}^{\prime}\) \(\left(\mathrm{g}^2+\mathrm{f}^2-6\right)=4\left(\mathrm{~g}^2+\mathrm{f}^2+3 \mathrm{~g}+3 \mathrm{f}\right)\) \(3 \mathrm{~g}^2+3 \mathrm{f}^2+12 \mathrm{~g}+12 \mathrm{f}+6=0\) \(\text { Dividing by } 3 \text { on both sides, }\) \(\mathrm{g}^2+\mathrm{f}^2+4 \mathrm{~g}+4 \mathrm{f}+2=0\)
AP EAMCET-22.09.2020
Conic Section
119882
If the lengths of the tangents drawn from \(P\) to the circles \(x^2+y^2-2 x+4 y-20=0\) and \(x^2+y^2-2 x-8 y+1=0\) are in the ratio 2:1, then the locus of \(P\) is
D We know that, Length of tangent drawn from \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to the circle \(x^2+y^2+2 g x+2 f y+c=0\) is \(\sqrt{x_1^2+y_1^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\) Let, \(\quad \mathrm{P}(\mathrm{h}, \mathrm{k})\), \(\therefore\) According to the question \(\frac{\sqrt{\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20}}{\sqrt{\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}-8 \mathrm{k}+1}}=\frac{2}{1}\) \(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20=4\left(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}-8 \mathrm{k}+1\right)\) \(\mathrm{h}^2+\mathrm{k}^2-2 \mathrm{~h}+4 \mathrm{k}-20=4 \mathrm{~h}^2+4 \mathrm{k}^2-8 \mathrm{~h}-32 \mathrm{k}+4\) \(3 \mathrm{~h}^2+3 \mathrm{k}^2-6 \mathrm{~h}-36 \mathrm{k}+24=0\) \(\mathrm{~h}^2+\mathrm{k}^2-2 \mathrm{~h}-12 \mathrm{k}+8=0\) \(\therefore \text { Locus of the point p is } \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-12 \mathrm{y}+8=0\)
AP EAMCET-22.04.2018
Conic Section
119883
Two tangents to the circle \(x^2+y^2=4\) at the points \(A\) and \(B\) meet at \(M(-4,0)\). The area of the quadrilateral MAOB, where \(O\) is the origin is
1 \(4 \sqrt{3}\) sq. units
2 \(2 \sqrt{3}\) sq. units
3 \(\sqrt{3}\) sq. units
4 \(3 \sqrt{3}\) sq. units
Explanation:
A \(\mathrm{x}^2+\mathrm{y}^2=4\) \(\mathrm{x}^2+\mathrm{y}^2=(2)^2\) \(\triangle \mathrm{OAM} \& \triangle \mathrm{OBM}\) \(\mathrm{OM}^2=\mathrm{AM}^2+\mathrm{OA}^2\) \((4)^2=(\mathrm{AM})^2+(2)^2\) \(\mathrm{AM}^2=16-4=12\) \(\mathrm{AM}=\sqrt{12}\) \(\text { Similarly } B M=\sqrt{12}\) \(\text { Area of OAMB }=2 \times \text { Area of } \triangle \mathrm{OAM}\) \(=2 \times \frac{1}{2} \times \mathrm{AM} \times \mathrm{OA}\) \(=\sqrt{12} \times 2=\sqrt{2 \times 2 \times 3} \times 2\) \(=2 \times 2 \sqrt{3}=4 \sqrt{3} \text { square unit }\)
