The point of contact of the tangent to the circle ${x^2} + {y^2} = 5$ at the point $(1, -2)$ which touches the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$, is
The point of contact of the tangent to the circle ${x^2} + {y^2} = 5$ at the point $(1, -2)$ which touches the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$, is
- A
$(2, -1)$
- B
$(3, -1)$
- C
$(4, -1)$
- D
$(5, -1)$
Similar Questions
If a line, $y=m x+c$ is a tangent to the circle, $(x-3)^{2}+y^{2}=1$ and it is perpendicular to a line $\mathrm{L}_{1},$ where $\mathrm{L}_{1}$ is the tangent to the circle, $\mathrm{x}^{2}+\mathrm{y}^{2}=1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),$ then
If a line, $y=m x+c$ is a tangent to the circle, $(x-3)^{2}+y^{2}=1$ and it is perpendicular to a line $\mathrm{L}_{1},$ where $\mathrm{L}_{1}$ is the tangent to the circle, $\mathrm{x}^{2}+\mathrm{y}^{2}=1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),$ then
- [JEE MAIN 2020]
$y - x + 3 = 0$ is the equation of normal at $\left( {3 + \frac{3}{{\sqrt 2 }},\frac{3}{{\sqrt 2 }}} \right)$ to which of the following circles
$y - x + 3 = 0$ is the equation of normal at $\left( {3 + \frac{3}{{\sqrt 2 }},\frac{3}{{\sqrt 2 }}} \right)$ to which of the following circles
The equation of the normal to the circle ${x^2} + {y^2} = 9$ at the point $\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ is
The equation of the normal to the circle ${x^2} + {y^2} = 9$ at the point $\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ is
The line $lx + my + n = 0$ will be a tangent to the circle ${x^2} + {y^2} = {a^2}$ if
The line $lx + my + n = 0$ will be a tangent to the circle ${x^2} + {y^2} = {a^2}$ if
Equation of a line through $(7, 4)$ and touching the circle, $x^2 + y^2 - 6x + 4y - 3 = 0$ is :
Equation of a line through $(7, 4)$ and touching the circle, $x^2 + y^2 - 6x + 4y - 3 = 0$ is :