If the shortest distance of the line 3 x+4 y+λ=0 from the circle x2+y2-4 x-6 y-3=0 is 1 then the sum of all possible value(s) of λ is

Question

If the shortest distance of the line 3 x+4 y+λ=0 from the circle x2+y2-4 x-6 y-3=0 is 1 then the sum of all possible value(s) of λ is (A) 43 (B) 7 (C) -36 (D) -43

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d892d8fd718a5bb153b71f7aecd58c5d-.png" /> ⇒ (|6+12+λ|/5)=5 ⇒ λ+18=-25,25 λ=-43,7 ∴ Sum =-36

Q. If the shortest distance of the line $3 x + 4 y + λ = 0$ from the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ is 1 then the sum of all possible value(s) of $λ$ is

43

7

-36

-43

$\Rightarrow \frac{∣6 + 12 + λ ∣}{5} = 5$
$\Rightarrow λ + 18 = - 25, 25$
$λ = - 43, 7$
$∴$ Sum $= - 36$

Q. If the shortest distance of the line 3x+4y+λ=0 from the circle x2+y2−4x−6y−3=0 is 1 then the sum of all possible value(s) of λ is

43

7

-36

-43

⇒5∣6+12+λ∣​=5 ⇒λ+18=−25,25 λ=−43,7 ∴ Sum =−36

Q. If the shortest distance of the line $3 x + 4 y + λ = 0$ from the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ is 1 then the sum of all possible value(s) of $λ$ is

$\Rightarrow \frac{∣6 + 12 + λ ∣}{5} = 5$
$\Rightarrow λ + 18 = - 25, 25$
$λ = - 43, 7$
$∴$ Sum $= - 36$