Let $P$ be a moving point on the circle $x^2 + y^2 - 6x - 8y + 21 = 0$. Then, the maximum distance of $P$ from the vertex of the parabola $x^2 + 6x + y + 13 = 0$ is equal to:

  • A
    $8$
  • B
    $10$
  • C
    $12$
  • D
    $9$

Explore More

Similar Questions

If the curve $x^{2}+2 y^{2}=2$ intersects the line $x + y =1$ at two points $P$ and $Q$,then the angle subtended by the line segment $PQ$ at the origin is ...... .

The equation of the common tangent to the curves $y^2 = 8x$ and $xy = -1$ is:

Difficult
View Solution

Consider an ellipse $E$,a hyperbola $H$,and a parabola $P$ such that each curve has the focus at $(2, 3)$ and the corresponding directrix is $x + y - 10 = 0$. If $(\alpha, \alpha_1)$,$(\beta, \beta_1)$,and $(\gamma, \gamma_1)$ are the nearest vertices of the ellipse,hyperbola,and parabola to the given directrix respectively,then:

Let $\lim_{x \to 2} \frac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$ for some $r, p \in R$. If the set of all possible values of $q$, such that the roots of the equation $rx^2 - px + q = 0$ lie in $(0, 2)$, be the interval $(\alpha, \beta]$, then $4(\alpha + \beta)$ equals :

Find the common tangent to the curves $x^2 + y^2 = 4$ and $2x^2 + y^2 = 2$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo