If the circles x2+y2+6 x+8 y+16=0 and x2+y2+2(3-√3) x+x+2(4-√6) y = k +6 √3+8 √6, k 0 touch internally at the point P(α, β), then (α+√3)2+(β+√6)2 is equal to

Question

If the circles x2+y2+6 x+8 y+16=0 and x2+y2+2(3-√3) x+x+2(4-√6) y = k +6 √3+8 √6, k >0 touch internally at the point P(α, β), then (α+√3)2+(β+√6)2 is equal to  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

C 1(-3,-4) r 1=√25-16=3 C 2=(-3+√3,-4+√6) r 2=√34+ k C 1 C 2=| r 1- r 2| C 1 C 2=√3+6=3 3=|3-√34+ k | ⇒ k =2 r 2=6 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/269fe07742e186c1c82189623e6b3087-.png" /> (α, β)=(-√3-3,-4-√6) (α+√3)2+(β+√6)2=9+16=25

Q. If the circles $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ and $x^{2} + y^{2} + 2 (3 - 3) x + x + 2 (4 - 6) y$ $= k + 63 + 86, k > 0$ touch internally at the point $P (α, β)$ , then $(α + 3)^{2} + (β + 6)^{2}$ is equal to _______

$C_{1} (- 3, - 4)$
$r_{1} = 25 - 16 = 3$
$C_{2} = (- 3 + 3, - 4 + 6)$
$r_{2} = 34 + k$
$C_{1} C_{2} = ∣ r_{1} - r_{2} ∣$
$C_{1} C_{2} = 3 + 6 = 3$
$3 = ∣3 - 34 + k ∣ \Rightarrow k = 2$
$r_{2} = 6$

$(α, β) = (- 3 - 3, - 4 - 6)$
$(α + 3)^{2} + (β + 6)^{2} = 9 + 16 = 25$

Q. If the circles x2+y2+6x+8y+16=0 and x2+y2+2(3−3​)x+x+2(4−6​)y =k+63​+86​,k>0 touch internally at the point P(α,β), then (α+3​)2+(β+6​)2 is equal to _______

C1​(−3,−4) r1​=25−16​=3 C2​=(−3+3​,−4+6​) r2​=34+k​ C1​C2​=∣r1​−r2​∣ C1​C2​=3+6​=3 3=∣3−34+k​∣⇒k=2 r2​=6 (α,β)=(−3​−3,−4−6​) (α+3​)2+(β+6​)2=9+16=25

Q. If the circles $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ and $x^{2} + y^{2} + 2 (3 - 3) x + x + 2 (4 - 6) y$ $= k + 63 + 86, k > 0$ touch internally at the point $P (α, β)$ , then $(α + 3)^{2} + (β + 6)^{2}$ is equal to _______

$C_{1} (- 3, - 4)$
$r_{1} = 25 - 16 = 3$
$C_{2} = (- 3 + 3, - 4 + 6)$
$r_{2} = 34 + k$
$C_{1} C_{2} = ∣ r_{1} - r_{2} ∣$
$C_{1} C_{2} = 3 + 6 = 3$
$3 = ∣3 - 34 + k ∣ \Rightarrow k = 2$
$r_{2} = 6$

$(α, β) = (- 3 - 3, - 4 - 6)$
$(α + 3)^{2} + (β + 6)^{2} = 9 + 16 = 25$