Let the equation of the circle,which touches | vedclass

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

Question

(D) The circle touches the $x$-axis at $(a, 0)$ and lies below the $x$-axis,so its center is $(a, -r)$ and its radius is $r$,where $r > 0$.The equatio

Vedclass · Accepted Answer

$(\alpha, \beta^2 - 4\gamma)$

Vedclass · Answer

(D) The circle touches the $x$-axis at $(a, 0)$ and lies below the $x$-axis,so its center is $(a, -r)$ and its radius is $r$,where $r > 0$.The equation of the circle is $(x - a)^2 + (y + r)^2 = r^2$,which simplifies to $x^2 - 2ax + a^2 + y^2 + 2ry + r^2 = r^2$,or $x^2 + y^2 - 2ax + 2ry + a^2 = 0$.Comparing this with $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$,we get $\alpha = 2a$,$\beta = 2r$,and $\gamma = a^2$.The circle cuts an intercept of length $b$ on the $y$-axis. Setting $x = 0$ in the equation,we get $y^2 + \beta y + \gamma = 0$. The roots are $y_1, y_2$,and $|y_1 - y_2| = b$.Using the property of roots,$|y_1 - y_2| = \sqrt{\beta^2 - 4\gamma} = b$,so $b^2 = \beta^2 - 4\gamma$.Thus,the ordered pair $(2a, b^2)$ is equal to $(\alpha, \beta^2 - 4\gamma)$.

Let the equation of the circle,which touches the $x$-axis at the point $(a, 0), a > 0$ and cuts off an intercept of length $b$ on the $y$-axis,be $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below the $x$-axis,then the ordered pair $(2a, b^2)$ is equal to:

Similar Questions

If $A(-2, 1)$,$B(0, -2)$,and $C(1, 2)$ are the vertices of a triangle $ABC$,then the perpendicular distance from its circumcentre to the side $BC$ is

$A$ circle with center at $(2,4)$ is such that the line $x+y+2=0$ cuts a chord of length $6$. The radius of the circle is

Consider a triangle $\Delta$ whose two sides lie on the $x$-axis and the line $x+y+1=0$. If the orthocenter of $\Delta$ is $(1,1)$,then the equation of the circle passing through the vertices of the triangle $\Delta$ is

$A$ point $P$ lies on the circle $x^{2}+y^{2}=169$. If $Q=(5, 12)$ and $R=(-12, 5)$, then the $\angle QPR$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module