The point of contact of the given circles {x^ | vedclass

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

Question

(b) ${x^2} + {y^2} - 6x - 6y + 10 = 0$&hellip;.$(i)$

${x^2} + {y^2} = 2$&hellip;.$(ii)$

$ \Rightarrow - 6x - 6y + 12 = 0$

or $x + y - 2 = 0$&

Vedclass · Accepted Answer

$(1, 1)$

Vedclass · Answer

(b) ${x^2} + {y^2} - 6x - 6y + 10 = 0$&hellip;.$(i)$

${x^2} + {y^2} = 2$&hellip;.$(ii)$

$ \Rightarrow - 6x - 6y + 12 = 0$

or $x + y - 2 = 0$&hellip;.$(iii)$

$ \Rightarrow {x^2} + {y^2} + 2xy = 4$ {from $(iii)$}

or $2xy = 2$ {from $(ii)$}

and $x - y = \sqrt {{{(x + y)}^2} - 4xy}$

$= \sqrt {4 - 4} = 0$

or $x = y$ and $x + y = 2$

$ \Rightarrow x = 1,\;y = 1$.

Trick : Required point must satisfy both the circles.

Obviously $(1, 1)$ satisfies both the equations simultaneously.

The point of contact of the given circles ${x^2} + {y^2} - 6x - 6y + 10 = 0$ and ${x^2} + {y^2} = 2$, is

Similar Questions

$P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the co-ordinate axes cut at right angles, then :

The range of values of $'a'$ such that the angle $\theta$ between the pair of tangents drawn from the point $(a, 0)$ to the circle $x^2 + y^2 = 1$ satisfies $\frac{\pi }{2} < \theta < \pi$ is :

Choose the correct statement about two circles whose equations are given below

$x^{2}+y^{2}-10 x-10 y+41=0$

$x^{2}+y^{2}-22 x-10 y+137=0$

If the circles ${x^2} + {y^2} - 9 = 0$ and ${x^2} + {y^2} + 2ax + 2y + 1 = 0$ touch each other, then $a =$

If $P$ and $Q$ are the points of intersection of the circles ${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$ and ${x^2} + {y^2} + 2x + 2y - {p^2} = 0$ then there is a circle passing through $P, Q$ and $(1, 1)$ for: