Let C: x^2+y^2=4 and C^{prime}: x^2+y^2-4 lam | vedclass

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(D) For two circles to intersect at two distinct points,the distance between their centers $d$ must satisfy $|r_1-r_2| < d < r_1+r_2$.For circle $C: x

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$6x^2+y^2=42$

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(D) For two circles to intersect at two distinct points,the distance between their centers $d$ must satisfy $|r_1-r_2| For circle $C: x^2+y^2=4$,center $C_1 = (0,0)$ and radius $r_1 = 2$.For circle $C^{\prime}: x^2+y^2-4\lambda x+9=0$,center $C_2 = (2\lambda, 0)$ and radius $r_2 = \sqrt{(2\lambda)^2 - 9} = \sqrt{4\lambda^2-9}$.For $r_2$ to be real,$4\lambda^2-9 \geq 0 \implies \lambda^2 \geq \frac{9}{4} \implies \lambda \in (-\infty, -\frac{3}{2}] \cup [\frac{3}{2}, \infty)$.The distance $d = |2\lambda - 0| = 2|\lambda|$.The condition $|r_1-r_2| Solving $2|\lambda| $2|\lambda| - 2  \frac{13}{8}$.Solving $|2 - \sqrt{4\lambda^2-9}| This is always true for the domain of $\lambda$ where $r_2$ is defined.Thus,$\lambda \in (-\infty, -\frac{13}{8}) \cup (\frac{13}{8}, \infty) = \mathbb{R} - [-\frac{13}{8}, \frac{13}{8}]$.Here $a = -\frac{13}{8}$ and $b = \frac{13}{8}$.The point is $(8(-\frac{13}{8})+12, 16(\frac{13}{8})-20) = (-13+12, 26-20) = (-1, 6)$.Checking the options: For $(-1, 6)$,$6(-1)^2 + (6)^2 = 6 + 36 = 42$. Thus,it lies on $6x^2+y^2=42$.

Let $C: x^2+y^2=4$ and $C^{\prime}: x^2+y^2-4 \lambda x+9=0$ be two circles. If the set of all values of $\lambda$ such that the circles $C$ and $C^{\prime}$ intersect at two distinct points is $\mathbb{R}-[a, b]$,then the point $(8a+12, 16b-20)$ lies on the curve:

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