If the curves x^2 - 6x + y^2 + 8 = 0 and x^2 | vedclass

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

Question

(B) The given equations of the circles are:$C_1: x^2 - 6x + y^2 + 8 = 0 \implies (x-3)^2 + y^2 = 1$. Center $C_1 = (3, 0)$,radius $r_1 = 1$.$C_2: x^2

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) The given equations of the circles are:$C_1: x^2 - 6x + y^2 + 8 = 0 \implies (x-3)^2 + y^2 = 1$. Center $C_1 = (3, 0)$,radius $r_1 = 1$.$C_2: x^2 + y^2 - 8y + 16 - k = 0 \implies x^2 + (y-4)^2 = k$. Center $C_2 = (0, 4)$,radius $r_2 = \sqrt{k}$.The distance between the centers $d = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{9 + 16} = 5$.For the circles to touch each other,the distance between centers must be equal to the sum or difference of the radii:Case $1$: $d = r_1 + r_2 \implies 5 = 1 + \sqrt{k} \implies \sqrt{k} = 4 \implies k = 16$.Case $2$: $d = |r_1 - r_2| \implies 5 = |1 - \sqrt{k}|$.This implies $1 - \sqrt{k} = 5$ (impossible as $k>0$) or $1 - \sqrt{k} = -5 \implies \sqrt{k} = 6 \implies k = 36$.Comparing the values of $k$,the largest value is $36$.

If the curves $x^2 - 6x + y^2 + 8 = 0$ and $x^2 - 8y + y^2 + 16 - k = 0$ $(k > 0)$ touch each other at a point,then the largest value of $k$ is

Similar Questions

$A$ tank is filled up to a height $h$ with a liquid and is placed on a platform of height $h$ from the ground. To get the maximum range $x_m$,a small hole is punched at a distance of $y$ from the free surface of the liquid. Then:

$A$ sample of a radioactive element has a mass of $10 \, g$ at an instant $t=0$. The approximate mass of this element in the sample after two mean lives is ........ $g$.

If in a $\triangle ABC$,$\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$,then $\angle C$ is equal to (in $^{\circ}$)

Bamboo and grasses elongate by the activity of ......

In which of the following is the $P - O - P$ bond present?

Vedclass Test Series

Exam Paper Generator

Online Exam Module