119675
If the equation of the circle having its centre in the second quadrant touches the coordinate axes and also the line \(\frac{x}{5}+\frac{y}{12}=1\) is \(x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0\), then \(\lambda=\)
1 3
2 10
3 15
4 -2
Explanation:
B Given, equation of Line \(\frac{x}{5}+\frac{y}{12}=1\) and equation of circle \(x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0,\) Let \(r\) be the radius of the circle. circle touches coordinate axes and centre lies in second quadrant and centre \(c=(-r, r)\) circle touch \(\frac{x}{5}+\frac{y}{12}=1 \Rightarrow\left|\frac{12(-r)+5(r)-60}{\sqrt{25+144}}\right| \Rightarrow r=\left|\frac{-7 r-60}{13}\right|\) \(7 r+60= \pm 13 r\) \(r=10\) Therefore equation of circle is, \((x+10)^2+(y-10)^2=10^2\) \(x^2+y^2+20 x-20 y+100=0\)So, \(\lambda=10\)
AP EAMCET-22.04.2019
Conic Section
119676
The equation of a circle passing through the point \((2,8)\), touching the lines \(4 x-3 y-24=0\) and \(4 x+3 y-42=0\) and having the \(x\) coordinate of its centre less than or equal to 8 is
1 \(x^2+y^2+2 x-8 y-8=0\)
2 \(x^2+y^2-4 x-6 y-12=0\)
3 \(x^2+y^2-4 x-10 y+4=0\)
4 \(x^2+y^2-6 x-4 y-24=0\)
Explanation:
B Let, O (h, k) be the centre of the circle. Since the circle passes through the point \((2,8)\) \(\therefore \text { Radius of the circle }=\sqrt{(\mathrm{h}-2)^2+(\mathrm{k}-8)^2}\) The circle touches the lines \(4 x-3 y-24=0\) and \(4 x+\) \(3 \mathrm{y}-42=0\) \(\therefore\left|\frac{4 h-3 k-24}{\sqrt{4^2+5^2}}\right|=\left|\frac{4 h+3 k-42}{\sqrt{4^2+5^2}}\right|\) Solving then, we get \(4 \mathrm{~h}-3 \mathrm{k}-24= \pm(4 \mathrm{k}+3 \mathrm{~h}-42)\) \(\therefore 6 \mathrm{k}=18\) or \(\mathrm{k}=3\) by taking the positive sign and \(8 \mathrm{~h}=66\) or \(\mathrm{h}=\frac{66}{8}=\frac{33}{4}\) by taking positive sign Given, \(|\mathrm{h}| \leq 8 \Rightarrow-8 \leq \mathrm{h} \leq 8\) \(\therefore \mathrm{h} \in \frac{33}{4}\) Put, \(\mathrm{k}=3\) in equation (i) and (ii) and equation, we get \((4 h-33)^2=(h-2)^2+25\) \(16 h^2-264 h+1089=25 h^2+725-100 h\) \(\text { on solving }\) \(\mathrm{h}=\frac{-164 \pm \sqrt{164^2-36 \mathrm{x}(-364)}}{2 \times 9}\) \(\mathrm{~h}=2, \frac{-182}{9}\) But, \(-8 \leq \mathrm{h} \leq 8 \therefore \mathrm{h}=2\) Now, \(\mathrm{r}^2=(\mathrm{h}-2)^2+(\mathrm{k}-8)^2 \Rightarrow(2-2)^2+(3-8)^2\) \(\mathrm{r}^2=25\) \(\mathrm{r}=5\) Hence, the equation of circle is \((x-2)^2+(y-3)^2=25\) \(x^2+y^2-4 x-6 y+4+9-25=0\) \(x^2+y^2-4 x-6 y-12=0\)
AP EAMCET-22.04.2019
Conic Section
119678
The largest value of a for which the circle \(\mathbf{x}^2+\mathrm{y}^2=\mathbf{a}^2\) lies completely in the interior of the parabola \(y^2=4 x+16\) is
1 \(4 \sqrt{3}\)
2 \(2 \sqrt{3}\)
3 \(2 \sqrt{6}\)
4 \(4 \sqrt{6}\)
Explanation:
B Given, equation of circle \(x^2+y^2=a^2\) falls entirely inside the parabola \(\mathrm{y}^2=4(\mathrm{x}+4)\) Substituting \(y^2=4(x+4)\) in \(x^2+y^2=a^2\) Gives, \(x^2+4 x+16-a^2=0\) Roots of the above quadratic should be real and equal. \(D=0\) \(16=4\left(16-a^2\right)\) \(a^2=12\) \(a=2 \sqrt{3}\)
AP EAMCET-22.09.2020
Conic Section
119679
The equation of circle of radius \(\sqrt{17} \mathrm{unit}\), with centre on the positive side of \(\mathrm{X}\)-axis and through the point \((0,1)\) is
1 \(x^2+y^2-8 x-1=0\)
2 \(x^2+y^2+8 x-1=0\)
3 \(x^2+y^2-9 y-1=0\)
4 \(2 x^2+2 y^2-3 x+2 y=4\)
Explanation:
A Given that, Radius \(=\sqrt{17}\) Centre on positive side of \(x\)-axis i.e., \((a, 0)\) \(\therefore\) Equation of circle in \((x-a)^2+y^2=\sqrt{17}\) Since, it passes through \((0,1)\) \(\therefore \mathrm{a}^2+1^2=17\) \(a^2=16\) \(\mathrm{a}=4\) Therefore equation of circle is, \((\mathrm{x}-4)^2+\mathrm{y}^2=17\) \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-1=0\)
119675
If the equation of the circle having its centre in the second quadrant touches the coordinate axes and also the line \(\frac{x}{5}+\frac{y}{12}=1\) is \(x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0\), then \(\lambda=\)
1 3
2 10
3 15
4 -2
Explanation:
B Given, equation of Line \(\frac{x}{5}+\frac{y}{12}=1\) and equation of circle \(x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0,\) Let \(r\) be the radius of the circle. circle touches coordinate axes and centre lies in second quadrant and centre \(c=(-r, r)\) circle touch \(\frac{x}{5}+\frac{y}{12}=1 \Rightarrow\left|\frac{12(-r)+5(r)-60}{\sqrt{25+144}}\right| \Rightarrow r=\left|\frac{-7 r-60}{13}\right|\) \(7 r+60= \pm 13 r\) \(r=10\) Therefore equation of circle is, \((x+10)^2+(y-10)^2=10^2\) \(x^2+y^2+20 x-20 y+100=0\)So, \(\lambda=10\)
AP EAMCET-22.04.2019
Conic Section
119676
The equation of a circle passing through the point \((2,8)\), touching the lines \(4 x-3 y-24=0\) and \(4 x+3 y-42=0\) and having the \(x\) coordinate of its centre less than or equal to 8 is
1 \(x^2+y^2+2 x-8 y-8=0\)
2 \(x^2+y^2-4 x-6 y-12=0\)
3 \(x^2+y^2-4 x-10 y+4=0\)
4 \(x^2+y^2-6 x-4 y-24=0\)
Explanation:
B Let, O (h, k) be the centre of the circle. Since the circle passes through the point \((2,8)\) \(\therefore \text { Radius of the circle }=\sqrt{(\mathrm{h}-2)^2+(\mathrm{k}-8)^2}\) The circle touches the lines \(4 x-3 y-24=0\) and \(4 x+\) \(3 \mathrm{y}-42=0\) \(\therefore\left|\frac{4 h-3 k-24}{\sqrt{4^2+5^2}}\right|=\left|\frac{4 h+3 k-42}{\sqrt{4^2+5^2}}\right|\) Solving then, we get \(4 \mathrm{~h}-3 \mathrm{k}-24= \pm(4 \mathrm{k}+3 \mathrm{~h}-42)\) \(\therefore 6 \mathrm{k}=18\) or \(\mathrm{k}=3\) by taking the positive sign and \(8 \mathrm{~h}=66\) or \(\mathrm{h}=\frac{66}{8}=\frac{33}{4}\) by taking positive sign Given, \(|\mathrm{h}| \leq 8 \Rightarrow-8 \leq \mathrm{h} \leq 8\) \(\therefore \mathrm{h} \in \frac{33}{4}\) Put, \(\mathrm{k}=3\) in equation (i) and (ii) and equation, we get \((4 h-33)^2=(h-2)^2+25\) \(16 h^2-264 h+1089=25 h^2+725-100 h\) \(\text { on solving }\) \(\mathrm{h}=\frac{-164 \pm \sqrt{164^2-36 \mathrm{x}(-364)}}{2 \times 9}\) \(\mathrm{~h}=2, \frac{-182}{9}\) But, \(-8 \leq \mathrm{h} \leq 8 \therefore \mathrm{h}=2\) Now, \(\mathrm{r}^2=(\mathrm{h}-2)^2+(\mathrm{k}-8)^2 \Rightarrow(2-2)^2+(3-8)^2\) \(\mathrm{r}^2=25\) \(\mathrm{r}=5\) Hence, the equation of circle is \((x-2)^2+(y-3)^2=25\) \(x^2+y^2-4 x-6 y+4+9-25=0\) \(x^2+y^2-4 x-6 y-12=0\)
AP EAMCET-22.04.2019
Conic Section
119678
The largest value of a for which the circle \(\mathbf{x}^2+\mathrm{y}^2=\mathbf{a}^2\) lies completely in the interior of the parabola \(y^2=4 x+16\) is
1 \(4 \sqrt{3}\)
2 \(2 \sqrt{3}\)
3 \(2 \sqrt{6}\)
4 \(4 \sqrt{6}\)
Explanation:
B Given, equation of circle \(x^2+y^2=a^2\) falls entirely inside the parabola \(\mathrm{y}^2=4(\mathrm{x}+4)\) Substituting \(y^2=4(x+4)\) in \(x^2+y^2=a^2\) Gives, \(x^2+4 x+16-a^2=0\) Roots of the above quadratic should be real and equal. \(D=0\) \(16=4\left(16-a^2\right)\) \(a^2=12\) \(a=2 \sqrt{3}\)
AP EAMCET-22.09.2020
Conic Section
119679
The equation of circle of radius \(\sqrt{17} \mathrm{unit}\), with centre on the positive side of \(\mathrm{X}\)-axis and through the point \((0,1)\) is
1 \(x^2+y^2-8 x-1=0\)
2 \(x^2+y^2+8 x-1=0\)
3 \(x^2+y^2-9 y-1=0\)
4 \(2 x^2+2 y^2-3 x+2 y=4\)
Explanation:
A Given that, Radius \(=\sqrt{17}\) Centre on positive side of \(x\)-axis i.e., \((a, 0)\) \(\therefore\) Equation of circle in \((x-a)^2+y^2=\sqrt{17}\) Since, it passes through \((0,1)\) \(\therefore \mathrm{a}^2+1^2=17\) \(a^2=16\) \(\mathrm{a}=4\) Therefore equation of circle is, \((\mathrm{x}-4)^2+\mathrm{y}^2=17\) \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-1=0\)
119675
If the equation of the circle having its centre in the second quadrant touches the coordinate axes and also the line \(\frac{x}{5}+\frac{y}{12}=1\) is \(x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0\), then \(\lambda=\)
1 3
2 10
3 15
4 -2
Explanation:
B Given, equation of Line \(\frac{x}{5}+\frac{y}{12}=1\) and equation of circle \(x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0,\) Let \(r\) be the radius of the circle. circle touches coordinate axes and centre lies in second quadrant and centre \(c=(-r, r)\) circle touch \(\frac{x}{5}+\frac{y}{12}=1 \Rightarrow\left|\frac{12(-r)+5(r)-60}{\sqrt{25+144}}\right| \Rightarrow r=\left|\frac{-7 r-60}{13}\right|\) \(7 r+60= \pm 13 r\) \(r=10\) Therefore equation of circle is, \((x+10)^2+(y-10)^2=10^2\) \(x^2+y^2+20 x-20 y+100=0\)So, \(\lambda=10\)
AP EAMCET-22.04.2019
Conic Section
119676
The equation of a circle passing through the point \((2,8)\), touching the lines \(4 x-3 y-24=0\) and \(4 x+3 y-42=0\) and having the \(x\) coordinate of its centre less than or equal to 8 is
1 \(x^2+y^2+2 x-8 y-8=0\)
2 \(x^2+y^2-4 x-6 y-12=0\)
3 \(x^2+y^2-4 x-10 y+4=0\)
4 \(x^2+y^2-6 x-4 y-24=0\)
Explanation:
B Let, O (h, k) be the centre of the circle. Since the circle passes through the point \((2,8)\) \(\therefore \text { Radius of the circle }=\sqrt{(\mathrm{h}-2)^2+(\mathrm{k}-8)^2}\) The circle touches the lines \(4 x-3 y-24=0\) and \(4 x+\) \(3 \mathrm{y}-42=0\) \(\therefore\left|\frac{4 h-3 k-24}{\sqrt{4^2+5^2}}\right|=\left|\frac{4 h+3 k-42}{\sqrt{4^2+5^2}}\right|\) Solving then, we get \(4 \mathrm{~h}-3 \mathrm{k}-24= \pm(4 \mathrm{k}+3 \mathrm{~h}-42)\) \(\therefore 6 \mathrm{k}=18\) or \(\mathrm{k}=3\) by taking the positive sign and \(8 \mathrm{~h}=66\) or \(\mathrm{h}=\frac{66}{8}=\frac{33}{4}\) by taking positive sign Given, \(|\mathrm{h}| \leq 8 \Rightarrow-8 \leq \mathrm{h} \leq 8\) \(\therefore \mathrm{h} \in \frac{33}{4}\) Put, \(\mathrm{k}=3\) in equation (i) and (ii) and equation, we get \((4 h-33)^2=(h-2)^2+25\) \(16 h^2-264 h+1089=25 h^2+725-100 h\) \(\text { on solving }\) \(\mathrm{h}=\frac{-164 \pm \sqrt{164^2-36 \mathrm{x}(-364)}}{2 \times 9}\) \(\mathrm{~h}=2, \frac{-182}{9}\) But, \(-8 \leq \mathrm{h} \leq 8 \therefore \mathrm{h}=2\) Now, \(\mathrm{r}^2=(\mathrm{h}-2)^2+(\mathrm{k}-8)^2 \Rightarrow(2-2)^2+(3-8)^2\) \(\mathrm{r}^2=25\) \(\mathrm{r}=5\) Hence, the equation of circle is \((x-2)^2+(y-3)^2=25\) \(x^2+y^2-4 x-6 y+4+9-25=0\) \(x^2+y^2-4 x-6 y-12=0\)
AP EAMCET-22.04.2019
Conic Section
119678
The largest value of a for which the circle \(\mathbf{x}^2+\mathrm{y}^2=\mathbf{a}^2\) lies completely in the interior of the parabola \(y^2=4 x+16\) is
1 \(4 \sqrt{3}\)
2 \(2 \sqrt{3}\)
3 \(2 \sqrt{6}\)
4 \(4 \sqrt{6}\)
Explanation:
B Given, equation of circle \(x^2+y^2=a^2\) falls entirely inside the parabola \(\mathrm{y}^2=4(\mathrm{x}+4)\) Substituting \(y^2=4(x+4)\) in \(x^2+y^2=a^2\) Gives, \(x^2+4 x+16-a^2=0\) Roots of the above quadratic should be real and equal. \(D=0\) \(16=4\left(16-a^2\right)\) \(a^2=12\) \(a=2 \sqrt{3}\)
AP EAMCET-22.09.2020
Conic Section
119679
The equation of circle of radius \(\sqrt{17} \mathrm{unit}\), with centre on the positive side of \(\mathrm{X}\)-axis and through the point \((0,1)\) is
1 \(x^2+y^2-8 x-1=0\)
2 \(x^2+y^2+8 x-1=0\)
3 \(x^2+y^2-9 y-1=0\)
4 \(2 x^2+2 y^2-3 x+2 y=4\)
Explanation:
A Given that, Radius \(=\sqrt{17}\) Centre on positive side of \(x\)-axis i.e., \((a, 0)\) \(\therefore\) Equation of circle in \((x-a)^2+y^2=\sqrt{17}\) Since, it passes through \((0,1)\) \(\therefore \mathrm{a}^2+1^2=17\) \(a^2=16\) \(\mathrm{a}=4\) Therefore equation of circle is, \((\mathrm{x}-4)^2+\mathrm{y}^2=17\) \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-1=0\)
119675
If the equation of the circle having its centre in the second quadrant touches the coordinate axes and also the line \(\frac{x}{5}+\frac{y}{12}=1\) is \(x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0\), then \(\lambda=\)
1 3
2 10
3 15
4 -2
Explanation:
B Given, equation of Line \(\frac{x}{5}+\frac{y}{12}=1\) and equation of circle \(x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0,\) Let \(r\) be the radius of the circle. circle touches coordinate axes and centre lies in second quadrant and centre \(c=(-r, r)\) circle touch \(\frac{x}{5}+\frac{y}{12}=1 \Rightarrow\left|\frac{12(-r)+5(r)-60}{\sqrt{25+144}}\right| \Rightarrow r=\left|\frac{-7 r-60}{13}\right|\) \(7 r+60= \pm 13 r\) \(r=10\) Therefore equation of circle is, \((x+10)^2+(y-10)^2=10^2\) \(x^2+y^2+20 x-20 y+100=0\)So, \(\lambda=10\)
AP EAMCET-22.04.2019
Conic Section
119676
The equation of a circle passing through the point \((2,8)\), touching the lines \(4 x-3 y-24=0\) and \(4 x+3 y-42=0\) and having the \(x\) coordinate of its centre less than or equal to 8 is
1 \(x^2+y^2+2 x-8 y-8=0\)
2 \(x^2+y^2-4 x-6 y-12=0\)
3 \(x^2+y^2-4 x-10 y+4=0\)
4 \(x^2+y^2-6 x-4 y-24=0\)
Explanation:
B Let, O (h, k) be the centre of the circle. Since the circle passes through the point \((2,8)\) \(\therefore \text { Radius of the circle }=\sqrt{(\mathrm{h}-2)^2+(\mathrm{k}-8)^2}\) The circle touches the lines \(4 x-3 y-24=0\) and \(4 x+\) \(3 \mathrm{y}-42=0\) \(\therefore\left|\frac{4 h-3 k-24}{\sqrt{4^2+5^2}}\right|=\left|\frac{4 h+3 k-42}{\sqrt{4^2+5^2}}\right|\) Solving then, we get \(4 \mathrm{~h}-3 \mathrm{k}-24= \pm(4 \mathrm{k}+3 \mathrm{~h}-42)\) \(\therefore 6 \mathrm{k}=18\) or \(\mathrm{k}=3\) by taking the positive sign and \(8 \mathrm{~h}=66\) or \(\mathrm{h}=\frac{66}{8}=\frac{33}{4}\) by taking positive sign Given, \(|\mathrm{h}| \leq 8 \Rightarrow-8 \leq \mathrm{h} \leq 8\) \(\therefore \mathrm{h} \in \frac{33}{4}\) Put, \(\mathrm{k}=3\) in equation (i) and (ii) and equation, we get \((4 h-33)^2=(h-2)^2+25\) \(16 h^2-264 h+1089=25 h^2+725-100 h\) \(\text { on solving }\) \(\mathrm{h}=\frac{-164 \pm \sqrt{164^2-36 \mathrm{x}(-364)}}{2 \times 9}\) \(\mathrm{~h}=2, \frac{-182}{9}\) But, \(-8 \leq \mathrm{h} \leq 8 \therefore \mathrm{h}=2\) Now, \(\mathrm{r}^2=(\mathrm{h}-2)^2+(\mathrm{k}-8)^2 \Rightarrow(2-2)^2+(3-8)^2\) \(\mathrm{r}^2=25\) \(\mathrm{r}=5\) Hence, the equation of circle is \((x-2)^2+(y-3)^2=25\) \(x^2+y^2-4 x-6 y+4+9-25=0\) \(x^2+y^2-4 x-6 y-12=0\)
AP EAMCET-22.04.2019
Conic Section
119678
The largest value of a for which the circle \(\mathbf{x}^2+\mathrm{y}^2=\mathbf{a}^2\) lies completely in the interior of the parabola \(y^2=4 x+16\) is
1 \(4 \sqrt{3}\)
2 \(2 \sqrt{3}\)
3 \(2 \sqrt{6}\)
4 \(4 \sqrt{6}\)
Explanation:
B Given, equation of circle \(x^2+y^2=a^2\) falls entirely inside the parabola \(\mathrm{y}^2=4(\mathrm{x}+4)\) Substituting \(y^2=4(x+4)\) in \(x^2+y^2=a^2\) Gives, \(x^2+4 x+16-a^2=0\) Roots of the above quadratic should be real and equal. \(D=0\) \(16=4\left(16-a^2\right)\) \(a^2=12\) \(a=2 \sqrt{3}\)
AP EAMCET-22.09.2020
Conic Section
119679
The equation of circle of radius \(\sqrt{17} \mathrm{unit}\), with centre on the positive side of \(\mathrm{X}\)-axis and through the point \((0,1)\) is
1 \(x^2+y^2-8 x-1=0\)
2 \(x^2+y^2+8 x-1=0\)
3 \(x^2+y^2-9 y-1=0\)
4 \(2 x^2+2 y^2-3 x+2 y=4\)
Explanation:
A Given that, Radius \(=\sqrt{17}\) Centre on positive side of \(x\)-axis i.e., \((a, 0)\) \(\therefore\) Equation of circle in \((x-a)^2+y^2=\sqrt{17}\) Since, it passes through \((0,1)\) \(\therefore \mathrm{a}^2+1^2=17\) \(a^2=16\) \(\mathrm{a}=4\) Therefore equation of circle is, \((\mathrm{x}-4)^2+\mathrm{y}^2=17\) \(\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-1=0\)
