The centre(s) of the circle(s) passing throug | vedclass

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

Question

(D) Let the equation of the circle passing through $(0, 0)$ and $(1, 0)$ be $x^2 + y^2 - x + \lambda y = 0$.This circle has its centre at $\left( \fra

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

Vedclass · Answer

(D) Let the equation of the circle passing through $(0, 0)$ and $(1, 0)$ be $x^2 + y^2 - x + \lambda y = 0$.This circle has its centre at $\left( \frac{1}{2}, -\frac{\lambda}{2} \right)$ and radius $r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\lambda}{2}\right)^2} = \sqrt{\frac{1 + \lambda^2}{4}} = \frac{\sqrt{1 + \lambda^2}}{2}$.The given circle is $x^2 + y^2 = 9$,which has centre $(0, 0)$ and radius $R = 3$.Since the two circles touch each other,the distance between their centres must be equal to the sum or difference of their radii: $d = |R \pm r|$.The distance between the centres is $d = \sqrt{\left(\frac{1}{2} - 0\right)^2 + \left(-\frac{\lambda}{2} - 0\right)^2} = \sqrt{\frac{1 + \lambda^2}{4}} = \frac{\sqrt{1 + \lambda^2}}{2}$.Thus,$\frac{\sqrt{1 + \lambda^2}}{2} = |3 \pm \frac{\sqrt{1 + \lambda^2}}{2}|$.Case $1$: $\frac{\sqrt{1 + \lambda^2}}{2} = 3 + \frac{\sqrt{1 + \lambda^2}}{2} \Rightarrow 3 = 0$ (Impossible).Case $2$: $\frac{\sqrt{1 + \lambda^2}}{2} = |3 - \frac{\sqrt{1 + \lambda^2}}{2}|$.If $3 - \frac{\sqrt{1 + \lambda^2}}{2} = \frac{\sqrt{1 + \lambda^2}}{2}$,then $\sqrt{1 + \lambda^2} = 3$ $\Rightarrow 1 + \lambda^2 = 9$ $\Rightarrow \lambda^2 = 8$ $\Rightarrow \lambda = \pm 2\sqrt{2}$.If $3 - \frac{\sqrt{1 + \lambda^2}}{2} = -\frac{\sqrt{1 + \lambda^2}}{2}$,then $3 = 0$ (Impossible).Thus,the centres are $\left( \frac{1}{2}, -\frac{\lambda}{2} \right) = \left( \frac{1}{2}, \mp \sqrt{2} \right)$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

The centre$(s)$ of the circle$(s)$ passing through the points $(0, 0)$ and $(1, 0)$ and touching the circle $x^2 + y^2 = 9$ is/are:

Similar Questions

Let $L_1$ be a line passing through the origin and $L_2$ be the line $x + y = 1$. If the intercepts made by the circle $x^{2} + y^{2} - x + 3y = 0$ on $L_1$ and $L_2$ are equal,then which of the following equations represents $L_1$?

The angle at which the circles $(x - 1)^2 + y^2 = 10$ and $x^2 + (y - 2)^2 = 5$ intersect is

The value of $c$ for which the set,$\{(x, y) | x^2 + y^2 + 2x \le 1 \} \cap \{(x, y) | x - y + c \ge 0\}$ contains only one point in common is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module