Find the equation of the circle passing throu | vedclass

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

Question

(A) Let the equation of the required circle be $(x - h)^{2} + (y - k)^{2} = r^{2}$.Since the circle passes through points $(2, 3)$ and $(-1, 1)$:$(2 -

Vedclass · Accepted Answer

$x^{2} + y^{2} - 7x + 5y - 14 = 0$

Vedclass · Answer

(A) Let the equation of the required circle be $(x - h)^{2} + (y - k)^{2} = r^{2}$.Since the circle passes through points $(2, 3)$ and $(-1, 1)$:$(2 - h)^{2} + (3 - k)^{2} = r^{2}$ $(1)$$(-1 - h)^{2} + (1 - k)^{2} = r^{2}$ $(2)$Since the centre $(h, k)$ of the circle lies on the line $x - 3y - 11 = 0$:$h - 3k = 11$ $(3)$From equations $(1)$ and $(2)$,we obtain:$(2 - h)^{2} + (3 - k)^{2} = (-1 - h)^{2} + (1 - k)^{2}$$4 - 4h + h^{2} + 9 - 6k + k^{2} = 1 + 2h + h^{2} + 1 - 2k + k^{2}$$13 - 4h - 6k = 2 + 2h - 2k$$6h + 4k = 11$ $(4)$Solving equations $(3)$ and $(4)$ for $h$ and $k$:Multiply $(3)$ by $2$: $2h - 6k = 22$Multiply $(4)$ by $3$: $18h + 12k = 33$Solving the system,we get $h = \frac{7}{2}$ and $k = -\frac{5}{2}$.Substituting $h$ and $k$ into $(1)$:$(2 - \frac{7}{2})^{2} + (3 + \frac{5}{2})^{2} = r^{2}$$(-\frac{3}{2})^{2} + (\frac{11}{2})^{2} = r^{2}$ $\Rightarrow \frac{9}{4} + \frac{121}{4} = r^{2}$ $\Rightarrow r^{2} = \frac{130}{4}$.The equation is $(x - \frac{7}{2})^{2} + (y + \frac{5}{2})^{2} = \frac{130}{4}$.$x^{2} - 7x + \frac{49}{4} + y^{2} + 5y + \frac{25}{4} = \frac{130}{4}$.$x^{2} + y^{2} - 7x + 5y + \frac{74}{4} - \frac{130}{4} = 0$.$x^{2} + y^{2} - 7x + 5y - \frac{56}{4} = 0$.$x^{2} + y^{2} - 7x + 5y - 14 = 0$.

Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre is on the line $x - 3y - 11 = 0$.

Similar Questions

The sides of a rectangle are given by $x = \pm a$ and $y = \pm b$. Then the equation of the circle passing through the vertices of the rectangle is

The four distinct points $(0,0), (2,0), (0,-2)$ and $(k,-2)$ are concyclic,if $k$ is equal to

If the equation of the circle lying in the first quadrant,touching both the coordinate axes and the line $\frac{x}{3}+\frac{y}{4}=1$ is $(x-c)^2+(y-c)^2=c^2$,then $c=$

$ABC$ is a triangle in which angle $C$ is a right angle. If the coordinates of $A$ and $B$ are $(-3, 4)$ and $(3, -4)$ respectively,then the equation of the circumcircle of triangle $ABC$ is

The centre of the circle $(x - 3)^2 + (y - 4)^2 = 5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module