The straight line x + 2y = 1 meets the coordi | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Equation of circle $x\left( {x - 1} \right) + \left( {y - \frac{1}{2}} \right)y = 0$

${x^2} + {y^2} - x - \frac{y}{2} = 0$

Equation of tang

Vedclass · Accepted Answer

$\frac {\sqrt 5}{2}$

Vedclass · Answer

Equation of circle $x\left( {x - 1} ight) + \left( {y - \frac{1}{2}} ight)y = 0$

${x^2} + {y^2} - x - \frac{y}{2} = 0$

Equation of tangent at $\left( {0,0} ight)$

$x.0 + y.0 - \frac{{x + 0}}{2} - \frac{{y + 0}}{{2 	imes 2}} = 0$

$2x + y = 0\,\,\,\,\,\,\,\,.......\left( 1 ight)$

Sum of distance of $A$ and $B$ from line $(i)$ is

$\frac{2}{{\sqrt 5 }} + \frac{{\frac{1}{2}}}{{\sqrt 5 }} = \frac{5}{{2\sqrt 5 }} = \frac{{\sqrt 5 }}{2}$

The straight line $x + 2y = 1$ meets the coordinate axes at $A$ and $B$. A circle is drawn through $A, B$ and the origin. Then the sum of perpendicular distances from $A$ and $B$ on the tangent to the circle at the origin is

Similar Questions

The equations of the tangents to the circle ${x^2} + {y^2} = 13$ at the points whose abscissa is $2$, are

If the straight line $4x + 3y + \lambda = 0$ touches the circle $2({x^2} + {y^2}) = 5$, then $\lambda $ is

The length of tangent from the point $(5, 1)$ to the circle ${x^2} + {y^2} + 6x - 4y - 3 = 0$, is

Let a circle $C$ touch the lines $L_{1}: 4 x-3 y+K_{1}$ $=0$ and $L _{2}: 4 x -3 y + K _{2}=0, K _{1}, K _{2} \in R$. If a line passing through the centre of the circle $C$ intersects $L _{1}$ at $(-1,2)$ and $L _{2}$ at $(3,-6)$, then the equation of the circle $C$ is

The equation of circle which touches the axes of coordinates and the line $\frac{x}{3} + \frac{y}{4} = 1$ and whose centre lies in the first quadrant is ${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$, where $c$ is