119659
The line \(x-2=0\) cuts the circle \(x^2+y^2-8 x-2 y+8=0\) at \(A\) and \(B\). The equation of the circle passing through the points \(A\) and \(B\) and having least radius is
1 \(x^2+y^2-4 x+2 y-1=0\)
2 \(x^2+y^2-4 x-2 y=0\)
3 \(x^2+y^2-4 x-2 y+1=0\)
4 \(x^2+y^2-4 x+4 y=0\)
Explanation:
B Equation of circle passing through \(\text { intersection of circle } S_1 \text { and line } L \text { is }\) \(\quad S_1+\lambda l=0\) \(\text { Equation of circle }=\left(x^2+y^2-8 x-2 y+8\right)+\lambda(x-2)=0\) \(x^2+y^2-8 x-2 y+8+\lambda x-2 \lambda=0\) \(x^2+y^2+x(-8+\lambda)-2 y+8-2 \lambda=0\) \(x^2+y^2+2 g x+2 f y+c=0\) \(\text { centre }(-g,-f)=[2 g=-8+\lambda][2 f=-2]\) \(\quad g=\frac{-8+\lambda}{2} \text { and } f=-1\) \(\quad(-g,-f)=\left(\frac{-\lambda+8}{2}, 1\right)\) \(\text { Centre lies on line L because the radius is least so }\) \(\text { The line is diameter }\) \(\quad \frac{x-2=0)}{2}=2\) \(\quad-\lambda+8=4\) \(\quad-\lambda=-4\) \(\lambda=4\) \(x^2+y^2+x(-8+4)-2 y+8-2(4)=0\) \(x^2+y^2-4 x-2 y+8-8=0\) \(x^2+y^2-4 x-2 y=0\) Centre lies on line \(\mathrm{L}\) because the radius is least so The line is diameter \((x-2=0)\) \(\frac{-\lambda+8}{2}=2\) \(-\lambda+8=4\) \(-\lambda=-4\) \(\lambda=4\) \(x^2+y^2+x(-8+4)-2 y+8-2(4)=0\) \(x^2+y^2-4 x-2 y+8-8=0\) \(x^2+y^2-4 x-2 y=0\)
AP EAMCET-22.04.2018
Conic Section
119660
The radius of a circle whose center lies in the fourth quadrant and touches each of the three lines \(x=0, y=0\) and \(3 x+4 y-12=0\) is \(\qquad\) \((\mathrm{x}-2=0)\) \(\frac{-\lambda+8}{2}=2\) \(-\lambda+8=4\) \(-\lambda=-4\) \(\lambda=4\) units.
119661
If the circle \(x^2+y^2-4 x-8 y-5=0\) intersects the line \(3 x-4 y-m=0\) in two distinct points then the number of integral values of ' \(\mathrm{m}\) ' is \(\qquad\)
1 52
2 51
3 50
4 49
Explanation:
D Given equation of circle, \(x^2+y^2-4 x-8 y-5=0\) \((x-2)^2+(y-4)^2-4-16-5=0\) \((x-2)^2+(y-4)^2=5^2\) \(\text { Centre of circle }=(2,4)\) \(\text { Radius of circle }=5\) \(\therefore 3 \mathrm{x}-4 \mathrm{y}=\mathrm{m}\) at two distinct points we know that, \(\left|\frac{6-16-\mathrm{m}}{\sqrt{3^2+4^2}}\right|\lt 5\) \(\left|\frac{\mathrm{m}+10}{5}\right|\lt 5\) \(|\mathrm{~m}+10|\lt 25\) \(-25\lt \mathrm{m}+10\lt 25\) \(-25-10\lt \mathrm{m}\lt 25-10\) \(-35\lt \mathrm{m}\lt 15\)Hence, the number of integral value of ' \(\mathrm{m}\) ' is 49.
AP EAPCET-25.08.2021
Conic Section
119662
The equation of a circle with center at \((-2,3)\) and circumference of \(4 \pi\) units is
1 \(x^2+y^2+4 x-6 y-9=0\)
2 \(x^2+y^2+4 x-6 y+9=0\)
3 \(x^2+y^2+4 x-6 y-3=0\)
4 \(x^2+y^2-4 x+6 y-9=0\)
Explanation:
B Circumference of circle is \(4 \pi\) \(2 \pi r=4 \pi\) \(r=2\) Equation of circle whose centre is \((-2,3)\) and radius 2 is , \((x+2)^2+(y-3)^2=(2)^2\) \(x^2+4+4 x+y^2+9-6 y=4\) \(x^2+y^2+4 x-6 y+9=0\)
119659
The line \(x-2=0\) cuts the circle \(x^2+y^2-8 x-2 y+8=0\) at \(A\) and \(B\). The equation of the circle passing through the points \(A\) and \(B\) and having least radius is
1 \(x^2+y^2-4 x+2 y-1=0\)
2 \(x^2+y^2-4 x-2 y=0\)
3 \(x^2+y^2-4 x-2 y+1=0\)
4 \(x^2+y^2-4 x+4 y=0\)
Explanation:
B Equation of circle passing through \(\text { intersection of circle } S_1 \text { and line } L \text { is }\) \(\quad S_1+\lambda l=0\) \(\text { Equation of circle }=\left(x^2+y^2-8 x-2 y+8\right)+\lambda(x-2)=0\) \(x^2+y^2-8 x-2 y+8+\lambda x-2 \lambda=0\) \(x^2+y^2+x(-8+\lambda)-2 y+8-2 \lambda=0\) \(x^2+y^2+2 g x+2 f y+c=0\) \(\text { centre }(-g,-f)=[2 g=-8+\lambda][2 f=-2]\) \(\quad g=\frac{-8+\lambda}{2} \text { and } f=-1\) \(\quad(-g,-f)=\left(\frac{-\lambda+8}{2}, 1\right)\) \(\text { Centre lies on line L because the radius is least so }\) \(\text { The line is diameter }\) \(\quad \frac{x-2=0)}{2}=2\) \(\quad-\lambda+8=4\) \(\quad-\lambda=-4\) \(\lambda=4\) \(x^2+y^2+x(-8+4)-2 y+8-2(4)=0\) \(x^2+y^2-4 x-2 y+8-8=0\) \(x^2+y^2-4 x-2 y=0\) Centre lies on line \(\mathrm{L}\) because the radius is least so The line is diameter \((x-2=0)\) \(\frac{-\lambda+8}{2}=2\) \(-\lambda+8=4\) \(-\lambda=-4\) \(\lambda=4\) \(x^2+y^2+x(-8+4)-2 y+8-2(4)=0\) \(x^2+y^2-4 x-2 y+8-8=0\) \(x^2+y^2-4 x-2 y=0\)
AP EAMCET-22.04.2018
Conic Section
119660
The radius of a circle whose center lies in the fourth quadrant and touches each of the three lines \(x=0, y=0\) and \(3 x+4 y-12=0\) is \(\qquad\) \((\mathrm{x}-2=0)\) \(\frac{-\lambda+8}{2}=2\) \(-\lambda+8=4\) \(-\lambda=-4\) \(\lambda=4\) units.
119661
If the circle \(x^2+y^2-4 x-8 y-5=0\) intersects the line \(3 x-4 y-m=0\) in two distinct points then the number of integral values of ' \(\mathrm{m}\) ' is \(\qquad\)
1 52
2 51
3 50
4 49
Explanation:
D Given equation of circle, \(x^2+y^2-4 x-8 y-5=0\) \((x-2)^2+(y-4)^2-4-16-5=0\) \((x-2)^2+(y-4)^2=5^2\) \(\text { Centre of circle }=(2,4)\) \(\text { Radius of circle }=5\) \(\therefore 3 \mathrm{x}-4 \mathrm{y}=\mathrm{m}\) at two distinct points we know that, \(\left|\frac{6-16-\mathrm{m}}{\sqrt{3^2+4^2}}\right|\lt 5\) \(\left|\frac{\mathrm{m}+10}{5}\right|\lt 5\) \(|\mathrm{~m}+10|\lt 25\) \(-25\lt \mathrm{m}+10\lt 25\) \(-25-10\lt \mathrm{m}\lt 25-10\) \(-35\lt \mathrm{m}\lt 15\)Hence, the number of integral value of ' \(\mathrm{m}\) ' is 49.
AP EAPCET-25.08.2021
Conic Section
119662
The equation of a circle with center at \((-2,3)\) and circumference of \(4 \pi\) units is
1 \(x^2+y^2+4 x-6 y-9=0\)
2 \(x^2+y^2+4 x-6 y+9=0\)
3 \(x^2+y^2+4 x-6 y-3=0\)
4 \(x^2+y^2-4 x+6 y-9=0\)
Explanation:
B Circumference of circle is \(4 \pi\) \(2 \pi r=4 \pi\) \(r=2\) Equation of circle whose centre is \((-2,3)\) and radius 2 is , \((x+2)^2+(y-3)^2=(2)^2\) \(x^2+4+4 x+y^2+9-6 y=4\) \(x^2+y^2+4 x-6 y+9=0\)
119659
The line \(x-2=0\) cuts the circle \(x^2+y^2-8 x-2 y+8=0\) at \(A\) and \(B\). The equation of the circle passing through the points \(A\) and \(B\) and having least radius is
1 \(x^2+y^2-4 x+2 y-1=0\)
2 \(x^2+y^2-4 x-2 y=0\)
3 \(x^2+y^2-4 x-2 y+1=0\)
4 \(x^2+y^2-4 x+4 y=0\)
Explanation:
B Equation of circle passing through \(\text { intersection of circle } S_1 \text { and line } L \text { is }\) \(\quad S_1+\lambda l=0\) \(\text { Equation of circle }=\left(x^2+y^2-8 x-2 y+8\right)+\lambda(x-2)=0\) \(x^2+y^2-8 x-2 y+8+\lambda x-2 \lambda=0\) \(x^2+y^2+x(-8+\lambda)-2 y+8-2 \lambda=0\) \(x^2+y^2+2 g x+2 f y+c=0\) \(\text { centre }(-g,-f)=[2 g=-8+\lambda][2 f=-2]\) \(\quad g=\frac{-8+\lambda}{2} \text { and } f=-1\) \(\quad(-g,-f)=\left(\frac{-\lambda+8}{2}, 1\right)\) \(\text { Centre lies on line L because the radius is least so }\) \(\text { The line is diameter }\) \(\quad \frac{x-2=0)}{2}=2\) \(\quad-\lambda+8=4\) \(\quad-\lambda=-4\) \(\lambda=4\) \(x^2+y^2+x(-8+4)-2 y+8-2(4)=0\) \(x^2+y^2-4 x-2 y+8-8=0\) \(x^2+y^2-4 x-2 y=0\) Centre lies on line \(\mathrm{L}\) because the radius is least so The line is diameter \((x-2=0)\) \(\frac{-\lambda+8}{2}=2\) \(-\lambda+8=4\) \(-\lambda=-4\) \(\lambda=4\) \(x^2+y^2+x(-8+4)-2 y+8-2(4)=0\) \(x^2+y^2-4 x-2 y+8-8=0\) \(x^2+y^2-4 x-2 y=0\)
AP EAMCET-22.04.2018
Conic Section
119660
The radius of a circle whose center lies in the fourth quadrant and touches each of the three lines \(x=0, y=0\) and \(3 x+4 y-12=0\) is \(\qquad\) \((\mathrm{x}-2=0)\) \(\frac{-\lambda+8}{2}=2\) \(-\lambda+8=4\) \(-\lambda=-4\) \(\lambda=4\) units.
119661
If the circle \(x^2+y^2-4 x-8 y-5=0\) intersects the line \(3 x-4 y-m=0\) in two distinct points then the number of integral values of ' \(\mathrm{m}\) ' is \(\qquad\)
1 52
2 51
3 50
4 49
Explanation:
D Given equation of circle, \(x^2+y^2-4 x-8 y-5=0\) \((x-2)^2+(y-4)^2-4-16-5=0\) \((x-2)^2+(y-4)^2=5^2\) \(\text { Centre of circle }=(2,4)\) \(\text { Radius of circle }=5\) \(\therefore 3 \mathrm{x}-4 \mathrm{y}=\mathrm{m}\) at two distinct points we know that, \(\left|\frac{6-16-\mathrm{m}}{\sqrt{3^2+4^2}}\right|\lt 5\) \(\left|\frac{\mathrm{m}+10}{5}\right|\lt 5\) \(|\mathrm{~m}+10|\lt 25\) \(-25\lt \mathrm{m}+10\lt 25\) \(-25-10\lt \mathrm{m}\lt 25-10\) \(-35\lt \mathrm{m}\lt 15\)Hence, the number of integral value of ' \(\mathrm{m}\) ' is 49.
AP EAPCET-25.08.2021
Conic Section
119662
The equation of a circle with center at \((-2,3)\) and circumference of \(4 \pi\) units is
1 \(x^2+y^2+4 x-6 y-9=0\)
2 \(x^2+y^2+4 x-6 y+9=0\)
3 \(x^2+y^2+4 x-6 y-3=0\)
4 \(x^2+y^2-4 x+6 y-9=0\)
Explanation:
B Circumference of circle is \(4 \pi\) \(2 \pi r=4 \pi\) \(r=2\) Equation of circle whose centre is \((-2,3)\) and radius 2 is , \((x+2)^2+(y-3)^2=(2)^2\) \(x^2+4+4 x+y^2+9-6 y=4\) \(x^2+y^2+4 x-6 y+9=0\)
119659
The line \(x-2=0\) cuts the circle \(x^2+y^2-8 x-2 y+8=0\) at \(A\) and \(B\). The equation of the circle passing through the points \(A\) and \(B\) and having least radius is
1 \(x^2+y^2-4 x+2 y-1=0\)
2 \(x^2+y^2-4 x-2 y=0\)
3 \(x^2+y^2-4 x-2 y+1=0\)
4 \(x^2+y^2-4 x+4 y=0\)
Explanation:
B Equation of circle passing through \(\text { intersection of circle } S_1 \text { and line } L \text { is }\) \(\quad S_1+\lambda l=0\) \(\text { Equation of circle }=\left(x^2+y^2-8 x-2 y+8\right)+\lambda(x-2)=0\) \(x^2+y^2-8 x-2 y+8+\lambda x-2 \lambda=0\) \(x^2+y^2+x(-8+\lambda)-2 y+8-2 \lambda=0\) \(x^2+y^2+2 g x+2 f y+c=0\) \(\text { centre }(-g,-f)=[2 g=-8+\lambda][2 f=-2]\) \(\quad g=\frac{-8+\lambda}{2} \text { and } f=-1\) \(\quad(-g,-f)=\left(\frac{-\lambda+8}{2}, 1\right)\) \(\text { Centre lies on line L because the radius is least so }\) \(\text { The line is diameter }\) \(\quad \frac{x-2=0)}{2}=2\) \(\quad-\lambda+8=4\) \(\quad-\lambda=-4\) \(\lambda=4\) \(x^2+y^2+x(-8+4)-2 y+8-2(4)=0\) \(x^2+y^2-4 x-2 y+8-8=0\) \(x^2+y^2-4 x-2 y=0\) Centre lies on line \(\mathrm{L}\) because the radius is least so The line is diameter \((x-2=0)\) \(\frac{-\lambda+8}{2}=2\) \(-\lambda+8=4\) \(-\lambda=-4\) \(\lambda=4\) \(x^2+y^2+x(-8+4)-2 y+8-2(4)=0\) \(x^2+y^2-4 x-2 y+8-8=0\) \(x^2+y^2-4 x-2 y=0\)
AP EAMCET-22.04.2018
Conic Section
119660
The radius of a circle whose center lies in the fourth quadrant and touches each of the three lines \(x=0, y=0\) and \(3 x+4 y-12=0\) is \(\qquad\) \((\mathrm{x}-2=0)\) \(\frac{-\lambda+8}{2}=2\) \(-\lambda+8=4\) \(-\lambda=-4\) \(\lambda=4\) units.
119661
If the circle \(x^2+y^2-4 x-8 y-5=0\) intersects the line \(3 x-4 y-m=0\) in two distinct points then the number of integral values of ' \(\mathrm{m}\) ' is \(\qquad\)
1 52
2 51
3 50
4 49
Explanation:
D Given equation of circle, \(x^2+y^2-4 x-8 y-5=0\) \((x-2)^2+(y-4)^2-4-16-5=0\) \((x-2)^2+(y-4)^2=5^2\) \(\text { Centre of circle }=(2,4)\) \(\text { Radius of circle }=5\) \(\therefore 3 \mathrm{x}-4 \mathrm{y}=\mathrm{m}\) at two distinct points we know that, \(\left|\frac{6-16-\mathrm{m}}{\sqrt{3^2+4^2}}\right|\lt 5\) \(\left|\frac{\mathrm{m}+10}{5}\right|\lt 5\) \(|\mathrm{~m}+10|\lt 25\) \(-25\lt \mathrm{m}+10\lt 25\) \(-25-10\lt \mathrm{m}\lt 25-10\) \(-35\lt \mathrm{m}\lt 15\)Hence, the number of integral value of ' \(\mathrm{m}\) ' is 49.
AP EAPCET-25.08.2021
Conic Section
119662
The equation of a circle with center at \((-2,3)\) and circumference of \(4 \pi\) units is
1 \(x^2+y^2+4 x-6 y-9=0\)
2 \(x^2+y^2+4 x-6 y+9=0\)
3 \(x^2+y^2+4 x-6 y-3=0\)
4 \(x^2+y^2-4 x+6 y-9=0\)
Explanation:
B Circumference of circle is \(4 \pi\) \(2 \pi r=4 \pi\) \(r=2\) Equation of circle whose centre is \((-2,3)\) and radius 2 is , \((x+2)^2+(y-3)^2=(2)^2\) \(x^2+4+4 x+y^2+9-6 y=4\) \(x^2+y^2+4 x-6 y+9=0\)
