119654
Let \(f(x, y)=0\) be the equation of a circle. If \(f\) \((0, \lambda)=0\) has equal roots \(\lambda=1,1\) and \(f(\lambda, 0)\) has roots \(\lambda=\frac{1}{2}, 2\), then the centre of the circle is
1 \(\left(1, \frac{1}{2}\right)\)
2 \(\left(\frac{5}{4}, 1\right)\)
3 \((5,4)\)
4 \(\left(\frac{1}{2}, 1\right)\)
Explanation:
B General equation of circle \(f(x, y)=(x-h)^2+(y-k)^2-a^2=0\) \(f(0, \lambda)=\mathrm{h}^2+(\lambda-\mathrm{k})^2-\mathrm{a}^2=0\) \(\lambda^2-2 \mathrm{k} \lambda+\mathrm{k}^2+\mathrm{h}^2-\mathrm{a}^2=0\) \(\text { Equation has equal roots }(1,1)\) \(\text { Sum of roots }=-\mathrm{b} / \mathrm{a}\) \(2=2 \mathrm{k}\) \(\mathrm{k}=1\) \(\text { Also } 4 \mathrm{k}^2-4 \mathrm{k}^2-4 \mathrm{~h}^2+\mathrm{a}^2=0\) \(4 \mathrm{~h}^2=\mathrm{a}^2\) \(f(\lambda, 0)=(\lambda-\mathrm{h})^2+(0-\mathrm{k})^2-\mathrm{a}^2=0\) \(\lambda^2-2 h \lambda+h^2+k^2-a^2=0\) \(\frac{5}{2}=2 \mathrm{~h}\) \(\mathrm{~h}=\frac{5}{4}\) \(\therefore \text { Centre }(\mathrm{h}, \mathrm{k})=\left(\frac{5}{4}, 1\right)\) Equation has roots \(\lambda=\frac{1}{2}\) and 2 Sum of roots \(=-\mathrm{b} / \mathrm{a}\) \(\frac{5}{2}=2 \mathrm{~h}\) \(\mathrm{~h}=\frac{5}{4}\)
AMU-2010
Conic Section
119655
The equation of the circle whose radius is 5 and which touches the circle \(x^2+y^2-2 x-4 y-20=0\) externally at the point \((5,5)\) is
1 \((x-9)^2+(y-8)^2=5^2\)
2 \((x-5)^2+(y-5)^2=5^2\)
3 \((x-0)^2+(y-0)^2=5^2\)
4 none of these
Explanation:
A Let equation of circle, \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=5^2\) It touches the circle- \(x^2+y^2-2 x-4 y-20=0\) Then, radius of circle \(\sqrt{1^2+2^2+20}=\sqrt{25}=5 \text { unit }\) Two circle touch each other externally \((5,5)\) is the midpoint of the line Segment joining \((\mathrm{h}, \mathrm{k})\) and \((1,2)\) \(5=\frac{\mathrm{h}+1}{2}\) \(\mathrm{~h}=9\) \(5=\frac{\mathrm{k}+2}{2}\) \(\mathrm{k}=8\) Equation of circle \((x-9)^2+(y-8)^2=5^2\)
AMU-2017
Conic Section
119656
The locus of the middle points of the chords of the circle \(x^2+y^2=a^2\) which subtend a right angle at the centre is
1 \(x^2+y^2=\frac{a^2}{2}\)
2 \(x^2+y^2=2 a^2\)
3 \(x^2+y^2=\frac{a^2}{4}\)
4 None of these
Explanation:
A Circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) having centre at origin and radius a Let \(A B\) the chord of the circle which subtend a right angle at the centre and ( \(\mathrm{h}, \mathrm{k})\) be the mid-point of the chord. Now, \(\triangle \mathrm{ACO}\), \(\sin 45^{\circ}=\frac{\mathrm{OC}}{\mathrm{OA}}\) \(\frac{1}{\sqrt{2}}=\frac{\mathrm{OC}}{\mathrm{a}}\) \(\mathrm{OC}=\frac{\mathrm{a}}{\sqrt{2}}\) \(\sqrt{(0-\mathrm{h})^2+(0-\mathrm{k})^2}=\frac{\mathrm{a}}{\sqrt{2}}\) \(\mathrm{h}^2+\mathrm{k}^2=\frac{\mathrm{a}^2}{2}\) Hence, locus of \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}^2+\mathrm{y}^2=\frac{\mathrm{a}^2}{2}\)
AMU-2015
Conic Section
119657
The set of values of \(k\) for which the circle \(C\) : \(4 x^2+4 y^2-12 x+8 y+k=0\) lies inside the fourth quadrant and the point \(\left(1,-\frac{1}{3}\right)\) lies on or inside the circle \(\mathbf{C}\) is :
119658
The equation of the circle which touches the \(x-\) axis and \(y\)-axis at the points \((1,0)\) and \((0,1)\) respectively is
1 \(x^2+y^2-4 \overline{y+3}=0\)
2 \(x^2+y^2-2 y+2=0\)
3 \(x^2+y^2-2 x-2 y+2=0\)
4 \(x^2+y^2-2 x-2 y+1=0\)
Explanation:
D : Equation of circle is, \(\left(\mathrm{x}-\mathrm{x}_1\right)^2+\left(\mathrm{y}-\mathrm{y}_1\right)^2=\mathrm{r}^2\) \((\mathrm{x}-1)^2+(\mathrm{y}-1)^2=1^2\) \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2+1-2 \mathrm{y}=1\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-2 \mathrm{y}+1=0\)
119654
Let \(f(x, y)=0\) be the equation of a circle. If \(f\) \((0, \lambda)=0\) has equal roots \(\lambda=1,1\) and \(f(\lambda, 0)\) has roots \(\lambda=\frac{1}{2}, 2\), then the centre of the circle is
1 \(\left(1, \frac{1}{2}\right)\)
2 \(\left(\frac{5}{4}, 1\right)\)
3 \((5,4)\)
4 \(\left(\frac{1}{2}, 1\right)\)
Explanation:
B General equation of circle \(f(x, y)=(x-h)^2+(y-k)^2-a^2=0\) \(f(0, \lambda)=\mathrm{h}^2+(\lambda-\mathrm{k})^2-\mathrm{a}^2=0\) \(\lambda^2-2 \mathrm{k} \lambda+\mathrm{k}^2+\mathrm{h}^2-\mathrm{a}^2=0\) \(\text { Equation has equal roots }(1,1)\) \(\text { Sum of roots }=-\mathrm{b} / \mathrm{a}\) \(2=2 \mathrm{k}\) \(\mathrm{k}=1\) \(\text { Also } 4 \mathrm{k}^2-4 \mathrm{k}^2-4 \mathrm{~h}^2+\mathrm{a}^2=0\) \(4 \mathrm{~h}^2=\mathrm{a}^2\) \(f(\lambda, 0)=(\lambda-\mathrm{h})^2+(0-\mathrm{k})^2-\mathrm{a}^2=0\) \(\lambda^2-2 h \lambda+h^2+k^2-a^2=0\) \(\frac{5}{2}=2 \mathrm{~h}\) \(\mathrm{~h}=\frac{5}{4}\) \(\therefore \text { Centre }(\mathrm{h}, \mathrm{k})=\left(\frac{5}{4}, 1\right)\) Equation has roots \(\lambda=\frac{1}{2}\) and 2 Sum of roots \(=-\mathrm{b} / \mathrm{a}\) \(\frac{5}{2}=2 \mathrm{~h}\) \(\mathrm{~h}=\frac{5}{4}\)
AMU-2010
Conic Section
119655
The equation of the circle whose radius is 5 and which touches the circle \(x^2+y^2-2 x-4 y-20=0\) externally at the point \((5,5)\) is
1 \((x-9)^2+(y-8)^2=5^2\)
2 \((x-5)^2+(y-5)^2=5^2\)
3 \((x-0)^2+(y-0)^2=5^2\)
4 none of these
Explanation:
A Let equation of circle, \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=5^2\) It touches the circle- \(x^2+y^2-2 x-4 y-20=0\) Then, radius of circle \(\sqrt{1^2+2^2+20}=\sqrt{25}=5 \text { unit }\) Two circle touch each other externally \((5,5)\) is the midpoint of the line Segment joining \((\mathrm{h}, \mathrm{k})\) and \((1,2)\) \(5=\frac{\mathrm{h}+1}{2}\) \(\mathrm{~h}=9\) \(5=\frac{\mathrm{k}+2}{2}\) \(\mathrm{k}=8\) Equation of circle \((x-9)^2+(y-8)^2=5^2\)
AMU-2017
Conic Section
119656
The locus of the middle points of the chords of the circle \(x^2+y^2=a^2\) which subtend a right angle at the centre is
1 \(x^2+y^2=\frac{a^2}{2}\)
2 \(x^2+y^2=2 a^2\)
3 \(x^2+y^2=\frac{a^2}{4}\)
4 None of these
Explanation:
A Circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) having centre at origin and radius a Let \(A B\) the chord of the circle which subtend a right angle at the centre and ( \(\mathrm{h}, \mathrm{k})\) be the mid-point of the chord. Now, \(\triangle \mathrm{ACO}\), \(\sin 45^{\circ}=\frac{\mathrm{OC}}{\mathrm{OA}}\) \(\frac{1}{\sqrt{2}}=\frac{\mathrm{OC}}{\mathrm{a}}\) \(\mathrm{OC}=\frac{\mathrm{a}}{\sqrt{2}}\) \(\sqrt{(0-\mathrm{h})^2+(0-\mathrm{k})^2}=\frac{\mathrm{a}}{\sqrt{2}}\) \(\mathrm{h}^2+\mathrm{k}^2=\frac{\mathrm{a}^2}{2}\) Hence, locus of \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}^2+\mathrm{y}^2=\frac{\mathrm{a}^2}{2}\)
AMU-2015
Conic Section
119657
The set of values of \(k\) for which the circle \(C\) : \(4 x^2+4 y^2-12 x+8 y+k=0\) lies inside the fourth quadrant and the point \(\left(1,-\frac{1}{3}\right)\) lies on or inside the circle \(\mathbf{C}\) is :
119658
The equation of the circle which touches the \(x-\) axis and \(y\)-axis at the points \((1,0)\) and \((0,1)\) respectively is
1 \(x^2+y^2-4 \overline{y+3}=0\)
2 \(x^2+y^2-2 y+2=0\)
3 \(x^2+y^2-2 x-2 y+2=0\)
4 \(x^2+y^2-2 x-2 y+1=0\)
Explanation:
D : Equation of circle is, \(\left(\mathrm{x}-\mathrm{x}_1\right)^2+\left(\mathrm{y}-\mathrm{y}_1\right)^2=\mathrm{r}^2\) \((\mathrm{x}-1)^2+(\mathrm{y}-1)^2=1^2\) \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2+1-2 \mathrm{y}=1\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-2 \mathrm{y}+1=0\)
119654
Let \(f(x, y)=0\) be the equation of a circle. If \(f\) \((0, \lambda)=0\) has equal roots \(\lambda=1,1\) and \(f(\lambda, 0)\) has roots \(\lambda=\frac{1}{2}, 2\), then the centre of the circle is
1 \(\left(1, \frac{1}{2}\right)\)
2 \(\left(\frac{5}{4}, 1\right)\)
3 \((5,4)\)
4 \(\left(\frac{1}{2}, 1\right)\)
Explanation:
B General equation of circle \(f(x, y)=(x-h)^2+(y-k)^2-a^2=0\) \(f(0, \lambda)=\mathrm{h}^2+(\lambda-\mathrm{k})^2-\mathrm{a}^2=0\) \(\lambda^2-2 \mathrm{k} \lambda+\mathrm{k}^2+\mathrm{h}^2-\mathrm{a}^2=0\) \(\text { Equation has equal roots }(1,1)\) \(\text { Sum of roots }=-\mathrm{b} / \mathrm{a}\) \(2=2 \mathrm{k}\) \(\mathrm{k}=1\) \(\text { Also } 4 \mathrm{k}^2-4 \mathrm{k}^2-4 \mathrm{~h}^2+\mathrm{a}^2=0\) \(4 \mathrm{~h}^2=\mathrm{a}^2\) \(f(\lambda, 0)=(\lambda-\mathrm{h})^2+(0-\mathrm{k})^2-\mathrm{a}^2=0\) \(\lambda^2-2 h \lambda+h^2+k^2-a^2=0\) \(\frac{5}{2}=2 \mathrm{~h}\) \(\mathrm{~h}=\frac{5}{4}\) \(\therefore \text { Centre }(\mathrm{h}, \mathrm{k})=\left(\frac{5}{4}, 1\right)\) Equation has roots \(\lambda=\frac{1}{2}\) and 2 Sum of roots \(=-\mathrm{b} / \mathrm{a}\) \(\frac{5}{2}=2 \mathrm{~h}\) \(\mathrm{~h}=\frac{5}{4}\)
AMU-2010
Conic Section
119655
The equation of the circle whose radius is 5 and which touches the circle \(x^2+y^2-2 x-4 y-20=0\) externally at the point \((5,5)\) is
1 \((x-9)^2+(y-8)^2=5^2\)
2 \((x-5)^2+(y-5)^2=5^2\)
3 \((x-0)^2+(y-0)^2=5^2\)
4 none of these
Explanation:
A Let equation of circle, \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=5^2\) It touches the circle- \(x^2+y^2-2 x-4 y-20=0\) Then, radius of circle \(\sqrt{1^2+2^2+20}=\sqrt{25}=5 \text { unit }\) Two circle touch each other externally \((5,5)\) is the midpoint of the line Segment joining \((\mathrm{h}, \mathrm{k})\) and \((1,2)\) \(5=\frac{\mathrm{h}+1}{2}\) \(\mathrm{~h}=9\) \(5=\frac{\mathrm{k}+2}{2}\) \(\mathrm{k}=8\) Equation of circle \((x-9)^2+(y-8)^2=5^2\)
AMU-2017
Conic Section
119656
The locus of the middle points of the chords of the circle \(x^2+y^2=a^2\) which subtend a right angle at the centre is
1 \(x^2+y^2=\frac{a^2}{2}\)
2 \(x^2+y^2=2 a^2\)
3 \(x^2+y^2=\frac{a^2}{4}\)
4 None of these
Explanation:
A Circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) having centre at origin and radius a Let \(A B\) the chord of the circle which subtend a right angle at the centre and ( \(\mathrm{h}, \mathrm{k})\) be the mid-point of the chord. Now, \(\triangle \mathrm{ACO}\), \(\sin 45^{\circ}=\frac{\mathrm{OC}}{\mathrm{OA}}\) \(\frac{1}{\sqrt{2}}=\frac{\mathrm{OC}}{\mathrm{a}}\) \(\mathrm{OC}=\frac{\mathrm{a}}{\sqrt{2}}\) \(\sqrt{(0-\mathrm{h})^2+(0-\mathrm{k})^2}=\frac{\mathrm{a}}{\sqrt{2}}\) \(\mathrm{h}^2+\mathrm{k}^2=\frac{\mathrm{a}^2}{2}\) Hence, locus of \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}^2+\mathrm{y}^2=\frac{\mathrm{a}^2}{2}\)
AMU-2015
Conic Section
119657
The set of values of \(k\) for which the circle \(C\) : \(4 x^2+4 y^2-12 x+8 y+k=0\) lies inside the fourth quadrant and the point \(\left(1,-\frac{1}{3}\right)\) lies on or inside the circle \(\mathbf{C}\) is :
119658
The equation of the circle which touches the \(x-\) axis and \(y\)-axis at the points \((1,0)\) and \((0,1)\) respectively is
1 \(x^2+y^2-4 \overline{y+3}=0\)
2 \(x^2+y^2-2 y+2=0\)
3 \(x^2+y^2-2 x-2 y+2=0\)
4 \(x^2+y^2-2 x-2 y+1=0\)
Explanation:
D : Equation of circle is, \(\left(\mathrm{x}-\mathrm{x}_1\right)^2+\left(\mathrm{y}-\mathrm{y}_1\right)^2=\mathrm{r}^2\) \((\mathrm{x}-1)^2+(\mathrm{y}-1)^2=1^2\) \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2+1-2 \mathrm{y}=1\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-2 \mathrm{y}+1=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119654
Let \(f(x, y)=0\) be the equation of a circle. If \(f\) \((0, \lambda)=0\) has equal roots \(\lambda=1,1\) and \(f(\lambda, 0)\) has roots \(\lambda=\frac{1}{2}, 2\), then the centre of the circle is
1 \(\left(1, \frac{1}{2}\right)\)
2 \(\left(\frac{5}{4}, 1\right)\)
3 \((5,4)\)
4 \(\left(\frac{1}{2}, 1\right)\)
Explanation:
B General equation of circle \(f(x, y)=(x-h)^2+(y-k)^2-a^2=0\) \(f(0, \lambda)=\mathrm{h}^2+(\lambda-\mathrm{k})^2-\mathrm{a}^2=0\) \(\lambda^2-2 \mathrm{k} \lambda+\mathrm{k}^2+\mathrm{h}^2-\mathrm{a}^2=0\) \(\text { Equation has equal roots }(1,1)\) \(\text { Sum of roots }=-\mathrm{b} / \mathrm{a}\) \(2=2 \mathrm{k}\) \(\mathrm{k}=1\) \(\text { Also } 4 \mathrm{k}^2-4 \mathrm{k}^2-4 \mathrm{~h}^2+\mathrm{a}^2=0\) \(4 \mathrm{~h}^2=\mathrm{a}^2\) \(f(\lambda, 0)=(\lambda-\mathrm{h})^2+(0-\mathrm{k})^2-\mathrm{a}^2=0\) \(\lambda^2-2 h \lambda+h^2+k^2-a^2=0\) \(\frac{5}{2}=2 \mathrm{~h}\) \(\mathrm{~h}=\frac{5}{4}\) \(\therefore \text { Centre }(\mathrm{h}, \mathrm{k})=\left(\frac{5}{4}, 1\right)\) Equation has roots \(\lambda=\frac{1}{2}\) and 2 Sum of roots \(=-\mathrm{b} / \mathrm{a}\) \(\frac{5}{2}=2 \mathrm{~h}\) \(\mathrm{~h}=\frac{5}{4}\)
AMU-2010
Conic Section
119655
The equation of the circle whose radius is 5 and which touches the circle \(x^2+y^2-2 x-4 y-20=0\) externally at the point \((5,5)\) is
1 \((x-9)^2+(y-8)^2=5^2\)
2 \((x-5)^2+(y-5)^2=5^2\)
3 \((x-0)^2+(y-0)^2=5^2\)
4 none of these
Explanation:
A Let equation of circle, \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=5^2\) It touches the circle- \(x^2+y^2-2 x-4 y-20=0\) Then, radius of circle \(\sqrt{1^2+2^2+20}=\sqrt{25}=5 \text { unit }\) Two circle touch each other externally \((5,5)\) is the midpoint of the line Segment joining \((\mathrm{h}, \mathrm{k})\) and \((1,2)\) \(5=\frac{\mathrm{h}+1}{2}\) \(\mathrm{~h}=9\) \(5=\frac{\mathrm{k}+2}{2}\) \(\mathrm{k}=8\) Equation of circle \((x-9)^2+(y-8)^2=5^2\)
AMU-2017
Conic Section
119656
The locus of the middle points of the chords of the circle \(x^2+y^2=a^2\) which subtend a right angle at the centre is
1 \(x^2+y^2=\frac{a^2}{2}\)
2 \(x^2+y^2=2 a^2\)
3 \(x^2+y^2=\frac{a^2}{4}\)
4 None of these
Explanation:
A Circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) having centre at origin and radius a Let \(A B\) the chord of the circle which subtend a right angle at the centre and ( \(\mathrm{h}, \mathrm{k})\) be the mid-point of the chord. Now, \(\triangle \mathrm{ACO}\), \(\sin 45^{\circ}=\frac{\mathrm{OC}}{\mathrm{OA}}\) \(\frac{1}{\sqrt{2}}=\frac{\mathrm{OC}}{\mathrm{a}}\) \(\mathrm{OC}=\frac{\mathrm{a}}{\sqrt{2}}\) \(\sqrt{(0-\mathrm{h})^2+(0-\mathrm{k})^2}=\frac{\mathrm{a}}{\sqrt{2}}\) \(\mathrm{h}^2+\mathrm{k}^2=\frac{\mathrm{a}^2}{2}\) Hence, locus of \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}^2+\mathrm{y}^2=\frac{\mathrm{a}^2}{2}\)
AMU-2015
Conic Section
119657
The set of values of \(k\) for which the circle \(C\) : \(4 x^2+4 y^2-12 x+8 y+k=0\) lies inside the fourth quadrant and the point \(\left(1,-\frac{1}{3}\right)\) lies on or inside the circle \(\mathbf{C}\) is :
119658
The equation of the circle which touches the \(x-\) axis and \(y\)-axis at the points \((1,0)\) and \((0,1)\) respectively is
1 \(x^2+y^2-4 \overline{y+3}=0\)
2 \(x^2+y^2-2 y+2=0\)
3 \(x^2+y^2-2 x-2 y+2=0\)
4 \(x^2+y^2-2 x-2 y+1=0\)
Explanation:
D : Equation of circle is, \(\left(\mathrm{x}-\mathrm{x}_1\right)^2+\left(\mathrm{y}-\mathrm{y}_1\right)^2=\mathrm{r}^2\) \((\mathrm{x}-1)^2+(\mathrm{y}-1)^2=1^2\) \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2+1-2 \mathrm{y}=1\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-2 \mathrm{y}+1=0\)
119654
Let \(f(x, y)=0\) be the equation of a circle. If \(f\) \((0, \lambda)=0\) has equal roots \(\lambda=1,1\) and \(f(\lambda, 0)\) has roots \(\lambda=\frac{1}{2}, 2\), then the centre of the circle is
1 \(\left(1, \frac{1}{2}\right)\)
2 \(\left(\frac{5}{4}, 1\right)\)
3 \((5,4)\)
4 \(\left(\frac{1}{2}, 1\right)\)
Explanation:
B General equation of circle \(f(x, y)=(x-h)^2+(y-k)^2-a^2=0\) \(f(0, \lambda)=\mathrm{h}^2+(\lambda-\mathrm{k})^2-\mathrm{a}^2=0\) \(\lambda^2-2 \mathrm{k} \lambda+\mathrm{k}^2+\mathrm{h}^2-\mathrm{a}^2=0\) \(\text { Equation has equal roots }(1,1)\) \(\text { Sum of roots }=-\mathrm{b} / \mathrm{a}\) \(2=2 \mathrm{k}\) \(\mathrm{k}=1\) \(\text { Also } 4 \mathrm{k}^2-4 \mathrm{k}^2-4 \mathrm{~h}^2+\mathrm{a}^2=0\) \(4 \mathrm{~h}^2=\mathrm{a}^2\) \(f(\lambda, 0)=(\lambda-\mathrm{h})^2+(0-\mathrm{k})^2-\mathrm{a}^2=0\) \(\lambda^2-2 h \lambda+h^2+k^2-a^2=0\) \(\frac{5}{2}=2 \mathrm{~h}\) \(\mathrm{~h}=\frac{5}{4}\) \(\therefore \text { Centre }(\mathrm{h}, \mathrm{k})=\left(\frac{5}{4}, 1\right)\) Equation has roots \(\lambda=\frac{1}{2}\) and 2 Sum of roots \(=-\mathrm{b} / \mathrm{a}\) \(\frac{5}{2}=2 \mathrm{~h}\) \(\mathrm{~h}=\frac{5}{4}\)
AMU-2010
Conic Section
119655
The equation of the circle whose radius is 5 and which touches the circle \(x^2+y^2-2 x-4 y-20=0\) externally at the point \((5,5)\) is
1 \((x-9)^2+(y-8)^2=5^2\)
2 \((x-5)^2+(y-5)^2=5^2\)
3 \((x-0)^2+(y-0)^2=5^2\)
4 none of these
Explanation:
A Let equation of circle, \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=5^2\) It touches the circle- \(x^2+y^2-2 x-4 y-20=0\) Then, radius of circle \(\sqrt{1^2+2^2+20}=\sqrt{25}=5 \text { unit }\) Two circle touch each other externally \((5,5)\) is the midpoint of the line Segment joining \((\mathrm{h}, \mathrm{k})\) and \((1,2)\) \(5=\frac{\mathrm{h}+1}{2}\) \(\mathrm{~h}=9\) \(5=\frac{\mathrm{k}+2}{2}\) \(\mathrm{k}=8\) Equation of circle \((x-9)^2+(y-8)^2=5^2\)
AMU-2017
Conic Section
119656
The locus of the middle points of the chords of the circle \(x^2+y^2=a^2\) which subtend a right angle at the centre is
1 \(x^2+y^2=\frac{a^2}{2}\)
2 \(x^2+y^2=2 a^2\)
3 \(x^2+y^2=\frac{a^2}{4}\)
4 None of these
Explanation:
A Circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) having centre at origin and radius a Let \(A B\) the chord of the circle which subtend a right angle at the centre and ( \(\mathrm{h}, \mathrm{k})\) be the mid-point of the chord. Now, \(\triangle \mathrm{ACO}\), \(\sin 45^{\circ}=\frac{\mathrm{OC}}{\mathrm{OA}}\) \(\frac{1}{\sqrt{2}}=\frac{\mathrm{OC}}{\mathrm{a}}\) \(\mathrm{OC}=\frac{\mathrm{a}}{\sqrt{2}}\) \(\sqrt{(0-\mathrm{h})^2+(0-\mathrm{k})^2}=\frac{\mathrm{a}}{\sqrt{2}}\) \(\mathrm{h}^2+\mathrm{k}^2=\frac{\mathrm{a}^2}{2}\) Hence, locus of \(\mathrm{C}(\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}^2+\mathrm{y}^2=\frac{\mathrm{a}^2}{2}\)
AMU-2015
Conic Section
119657
The set of values of \(k\) for which the circle \(C\) : \(4 x^2+4 y^2-12 x+8 y+k=0\) lies inside the fourth quadrant and the point \(\left(1,-\frac{1}{3}\right)\) lies on or inside the circle \(\mathbf{C}\) is :
119658
The equation of the circle which touches the \(x-\) axis and \(y\)-axis at the points \((1,0)\) and \((0,1)\) respectively is
1 \(x^2+y^2-4 \overline{y+3}=0\)
2 \(x^2+y^2-2 y+2=0\)
3 \(x^2+y^2-2 x-2 y+2=0\)
4 \(x^2+y^2-2 x-2 y+1=0\)
Explanation:
D : Equation of circle is, \(\left(\mathrm{x}-\mathrm{x}_1\right)^2+\left(\mathrm{y}-\mathrm{y}_1\right)^2=\mathrm{r}^2\) \((\mathrm{x}-1)^2+(\mathrm{y}-1)^2=1^2\) \(\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2+1-2 \mathrm{y}=1\) \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-2 \mathrm{y}+1=0\)
