The diameters of a circle are along 2x + y - | vedclass

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

Question

(C) The intersection point of the diameter lines is the center of the circle. Solving the equations $2x + y = 7$ and $x + 3y = 11$:From the first equa

Vedclass · Accepted Answer

$x^2 + y^2 - 4x - 6y - 12 = 0$

Vedclass · Answer

(C) The intersection point of the diameter lines is the center of the circle. Solving the equations $2x + y = 7$ and $x + 3y = 11$:From the first equation,$y = 7 - 2x$.Substituting into the second: $x + 3(7 - 2x) = 11$ $\Rightarrow x + 21 - 6x = 11$ $\Rightarrow -5x = -10$ $\Rightarrow x = 2$.Then $y = 7 - 2(2) = 3$. So,the center $(h, k) = (2, 3)$.The circle passes through $(5, 7)$,so the radius $r$ is the distance between $(2, 3)$ and $(5, 7)$:$r = \sqrt{(5 - 2)^2 + (7 - 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$:$(x - 2)^2 + (y - 3)^2 = 5^2$$x^2 - 4x + 4 + y^2 - 6y + 9 = 25$$x^2 + y^2 - 4x - 6y + 13 - 25 = 0$$x^2 + y^2 - 4x - 6y - 12 = 0$.

The diameters of a circle are along $2x + y - 7 = 0$ and $x + 3y - 11 = 0$. Then,the equation of this circle,which also passes through $(5, 7)$,is

Similar Questions

The abscissae of $A$ and $B$ are the roots of the equation $x^2 + 2ax - b^2 = 0$ and their ordinates are the roots of the equation $y^2 + 2py - q^2 = 0$. Find the equation of the circle with $AB$ as the diameter.

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

If the line $3x - 2y + 6 = 0$ meets the $X$-axis and $Y$-axis at points $A$ and $B$ respectively,then the equation of the circle with radius $AB$ and center at $A$ is

If $(1, 1), (-2, 2), (2, -2)$ are $3$ points on a circle $S$,then the perpendicular distance from the centre of the circle $S$ to the line $3x - 4y + 1 = 0$ is

Find the equation of the circle with center $(3, 1)$ and touching the line $8x - 15y + 25 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module