If a circle passing through (1, -2) has x - y | vedclass

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

Question

(D) The center of the circle is the intersection point of its diameters $x - y = 2$ and $2x + 3y = 14$. Solving $x - y = 2$ for $y$,we get $y = x - 2$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) The center of the circle is the intersection point of its diameters $x - y = 2$ and $2x + 3y = 14$. Solving $x - y = 2$ for $y$,we get $y = x - 2$. Substituting this into the second equation: $2x + 3(x - 2) = 14$. $2x + 3x - 6 = 14$ $\Rightarrow 5x = 20$ $\Rightarrow x = 4$. Then $y = 4 - 2 = 2$. So,the center of the circle is $(4, 2)$. The circle passes through $(1, -2)$. The radius $r$ is the distance between the center $(4, 2)$ and the point $(1, -2)$. $r = \sqrt{(4 - 1)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

If a circle passing through $(1, -2)$ has $x - y = 2$ and $2x + 3y = 14$ as its diameters,then the radius of the circle is

Similar Questions

$A$ circle is drawn with the $y$-axis as a tangent and its center at the point which is the reflection of $(3, 4)$ in the line $y = x$. The equation of the circle is:

$A$ circle has a radius of $3$ units and its centre lies on the line $y = x - 1$. If the circle passes through the point $(7, 3)$,what is its equation?

Circles are drawn through the point $(2,0)$ to cut intercepts of length $5$ units on the $X$-axis. If their centre lies in the first quadrant,then their equation is

$P$ and $Q$ are the points of trisection of the line segment joining the points $(3, -7)$ and $(-5, 3)$. If $PQ$ subtends a right angle at a variable point $R$,then the locus of $R$ is

The equation $ax^2 + 2y^2 + 2bxy + 2x - y + c = 0$ represents a circle passing through the origin,if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module