If length of tangents drawn from point (1,2) to circles {x^2} + {y^2} + x + y

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10-1.Circle and System of Circles

hard

If the length of the tangents drawn from the point $(1,2)$ to the circles ${x^2} + {y^2} + x + y - 4 = 0$ and $3{x^2} + 3{y^2} - x - y + k = 0$ be in the ratio $4 : 3$, then $k =$

$7/2$

$21/2$

$-21/ 4$

$7/4$

Solution

(c) Given $\frac{{{T_1}}}{{{T_2}}} = \frac{4}{3}$, where $T_1$ and $T_2$ are the length of tangents drawn to the given circle. $ \Rightarrow \frac{{\sqrt {1 + 4 + 1 + 2 – 4} }}{{\sqrt {{{(1)}^2} + {{(2)}^2} – \frac{1}{3} – \frac{2}{3} + \frac{k}{3}} }} = \frac{4}{3} \Rightarrow k = – \frac{{21}}{4}$.

Standard 11

Mathematics

A tangent to the circle ${x^2} + {y^2} = 5$at the point $(1,-2)$ the circle ${x^2} + {y^2} – 8x + 6y + 20 = 0$

medium

(IIT-1975)

A pair of tangents are drawn from the origin to the circle ${x^2} + {y^2} + 20(x + y) + 20 = 0$. The equation of the pair of tangents is

hard

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

hard

(JEE MAIN-2022)

Number of integral points interior to the circle $x^2 + y^2 = 10$ from which exactly one real tangent can be drawn to the curve $\sqrt {{{\left( {x + 5\sqrt 2 } \right)}^2} + {y^2}} \, – \sqrt {{{\left( {x – 5\sqrt 2 } \right)}^2} + {y^2}\,} \, = 10$ are (where integral point $(x, y)$ means $x, y \in I)$

normal

Which of the following lines is a tangent to the circle ${x^2} + {y^2} = 25$ for all values of $m$…..

medium

If the length of the tangents drawn from the point $(1,2)$ to the circles ${x^2} + {y^2} + x + y - 4 = 0$ and $3{x^2} + 3{y^2} - x - y + k = 0$ be in the ratio $4 : 3$, then $k =$

Solution

Similar Questions

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