The point of contact of the given circles x^2 | vedclass

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

Question

(B) Given equations of the circles are:$x^2 + y^2 - 6x - 6y + 10 = 0$ $(i)$$x^2 + y^2 = 2$ $(ii)$Subtracting equation $(ii)$ from $(i)$:$(x^2 + y^2 -

Vedclass · Accepted Answer

$(1, 1)$

Vedclass · Answer

(B) Given equations of the circles are:$x^2 + y^2 - 6x - 6y + 10 = 0$ $(i)$$x^2 + y^2 = 2$ $(ii)$Subtracting equation $(ii)$ from $(i)$:$(x^2 + y^2 - 6x - 6y + 10) - (x^2 + y^2) = 0 - 2$$-6x - 6y + 10 = -2$$-6x - 6y + 12 = 0$$x + y = 2$ $(iii)$From $(iii)$,$y = 2 - x$. Substituting this into $(ii)$:$x^2 + (2 - x)^2 = 2$$x^2 + 4 - 4x + x^2 = 2$$2x^2 - 4x + 2 = 0$$x^2 - 2x + 1 = 0$$(x - 1)^2 = 0$$x = 1$Substituting $x = 1$ into $x + y = 2$,we get $y = 1$.Thus,the point of contact is $(1, 1)$.

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

$(a, 0)$ and $(b, 0)$ are the centres of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is

If the circles $x^{2}+y^{2}+2x+2ky+6=0$ and $x^{2}+y^{2}+2ky+k=0$ intersect orthogonally,then $k$ is

The two circles $x^2 + y^2 - 2x - 3 = 0$ and $x^2 + y^2 - 4x - 6y - 8 = 0$ are such that:

The equation of the circle which passes through the origin,has its centre on the line $x + y = 4$ and cuts the circle ${x^2} + {y^2} - 4x + 2y + 4 = 0$ orthogonally,is

The circles $x^2+y^2-10x+16=0$ and $x^2+y^2=a^2$ intersect at two distinct points if

Vedclass Test Series

Exam Paper Generator

Online Exam Module