Consider a family of circles which are passin | vedclass

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

Question

Equation of circle $(x-h)^{2}+(y-k)^{2}=k^{2}$

It is passing through $(-1,1)$ then $(-1-h)^{2}+(1-k)^{2}=k^{2}$

$\Rightarrow h^{2}+2 h-2 k+2=0$

Vedclass · Accepted Answer

$k \ge \frac{1}{2}$

Vedclass · Answer

Equation of circle $(x-h)^{2}+(y-k)^{2}=k^{2}$

It is passing through $(-1,1)$ then $(-1-h)^{2}+(1-k)^{2}=k^{2}$

$\Rightarrow h^{2}+2 h-2 k+2=0$

$D \geq 0$

$2 k-1 \geq 0 \Rightarrow k \geq 1 / 2$

Consider a family of circles which are passing through the point $(- 1, 1)$ and are tangent to $x-$ axis. If $(h, k)$ are the coordinate of the centre of the circles, then the set of values of $k$ is given by the interval

Similar Questions

If the circles of same radius a and centers at $(2, 3)$ and $(5, 6)$ cut orthogonally, then $a =$

The number of common tangents to the circles ${x^2} + {y^2} - x = 0,\,{x^2} + {y^2} + x = 0$ is

Circles ${(x + a)^2} + {(y + b)^2} = {a^2}$ and ${(x + \alpha )^2}$ $ + {(y + \beta )^2} = $ ${\beta ^2}$ cut orthogonally, if

The condition that the circle ${(x - 3)^2} + {(y - 4)^2} = {r^2}$ lies entirely within the circle ${x^2} + {y^2} = {R^2},$ is

Let $C_i \equiv x^2 + y^2 = i^2 (i = 1,2,3)$ are three circles. If there are $4i$ points on circumference of circle $C_i$. If no three of all the points on three circles are collinear then number of triangles which can be formed using these points whose circumcentre does not lie on origin, is-