The equation of the circle passing through th | vedclass

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

Question

Point $(1,2)$ lies on the circle ${x^2} + {y^2} + 2x + 2y - 11 = 0\,\,\,$, because coordinates of point $(1,2)$ satisfy the equation

${x^2} +

Vedclass · Accepted Answer

$x^2 +y^2 + 2x + 2y - 11 = 0$

Vedclass · Answer

Point $(1,2)$ lies on the circle ${x^2} + {y^2} + 2x + 2y - 11 = 0\,\,\,$, because coordinates of point $(1,2)$ satisfy the equation

${x^2} + {y^2} + 2x + 2y - 11 = 0\,\,\,$

Now, ${x^2} + {y^2} - 4x + 6y - 21 = 0\,\,\,\,\,\,\,\,\,........\left( i ight)$

${x^2} + {y^2} + 2x + 2y - 11 = 0\,\,\,\,\,\,\,\,\,..........\left( {ii} ight)$

$3x + 4y + 5 = 0\,\,\,\,\,\,\,\,\,..........\left( {iii} ight)$

From $(i)$ and $(iii)$

${x^2} + {\left( { - \frac{{3x + 5}}{4}} ight)^2} - 4x - 6{\left( { - \frac{{3x + 5}}{4}} ight)^2} - 21 = 0$

$\begin{array}{l}
 \Rightarrow 16{x^2} + 9{x^2} + 30x + 25 - 64x\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 72x + 120 - 336 = 0
\end{array}$

$ \Rightarrow 25{x^2} + 38x - 191 = 0\,\,\,\,\,\,\,\,\,\,.......\left( {iv} ight)$

From $(ii)$ abd $(iii)$,

${x^2} + {\left( { - \frac{{3x + 5}}{4}} ight)^2} + 2x + 2{\left( { - \frac{{3x + 5}}{4}} ight)^2} - 11 = 0$

$\begin{array}{l}
 \Rightarrow 16{x^2} + 9{x^2} + 30x + 25 + 32x\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 24x - 40 - 176 = 0
\end{array}$

$ \Rightarrow 25{x^2} + 38x - 191 = 0\,\,\,\,\,\,\,\,\,\,.......\left( v ight)$

Thus we get the same equation from $(ii)$ and $(iii)$ as we get from equation $(i)$ and $(iii)$. Hence the point of intersction of $(ii)$ and $(iii)$ will be sameas the point of intersections of $(i)$ and $(iii)$, Therefore the circle $(ii)$ passing through the point of  inersection of circle $(i)$ and poin $(1,2)$ also as show in the figure.

Hence equation $(ii)$ i.e

${x^2} + {y^2} + 2x + 2y - 11 = 0\,\,\,$ is the equation of required circle.

The equation of the circle passing through the point $(1, 2)$ and through the points of intersection of $x^2 + y^2 - 4x - 6y - 21 = 0$ and $3x + 4y + 5 = 0$ is given by

Similar Questions

The value of $\lambda $, for which the circle ${x^2} + {y^2} + 2\lambda x + 6y + 1 = 0$, intersects the circle ${x^2} + {y^2} + 4x + 2y = 0$ orthogonally is

The number of integral values of $\lambda $ for which $x^2 + y^2 + \lambda x + (1 - \lambda )y + 5 = 0$ is the equation of a circle whose radius cannot exceed $5$ , is

The locus of centre of a circle passing through $(a, b)$ and cuts orthogonally to circle ${x^2} + {y^2} = {p^2}$, is

The number of common tangents to the circles ${x^2} + {y^2} = 1$and ${x^2} + {y^2} - 4x + 3 = 0$ is

Number of common tangents to the circles
$x^2 + y^2 -2x + 4y -4 = 0$ and
$x^2 + y^2 -8x -4y + 16 = 0 $ is-