A circle C touches the X-axis and makes an in | vedclass

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

Question

(B) Let the centre of the circle $C$ be $(h, k)$. Since the circle touches the $X$-axis,the radius $r = |k|$.Given the centre lies on $y=x+1$,we have

Vedclass · Accepted Answer

$x^2+y^2-26x-20y+19=0$

Vedclass · Answer

(B) Let the centre of the circle $C$ be $(h, k)$. Since the circle touches the $X$-axis,the radius $r = |k|$.Given the centre lies on $y=x+1$,we have $k = h+1$.The equation of the circle is $(x-h)^2 + (y-k)^2 = k^2$.This circle makes an intercept of $2$ units on the $Y$-axis. Setting $x=0$,we get $h^2 + (y-k)^2 = k^2$,which simplifies to $y^2 - 2ky + h^2 = 0$.The length of the intercept is $|y_1 - y_2| = 2\sqrt{k^2 - h^2} = 2$.Thus,$k^2 - h^2 = 1$.Substituting $k = h+1$,we get $(h+1)^2 - h^2 = 1$,which implies $h^2 + 2h + 1 - h^2 = 1$,so $2h = 0$,giving $h=0$.Then $k = 0+1 = 1$.The centre of circle $C$ is $(0, 1)$.We need to find a circle passing through $(0, 1)$.Checking option $A$: $0^2 + 1^2 - 2(0) - 4(1) + 1 = 1 - 4 + 1 = -2 
eq 0$.Checking option $D$: $0^2 + 1^2 + 2(0) - 4(1) + 1 = 1 - 4 + 1 = -2 
eq 0$.Checking option $B$: $0^2 + 1^2 - 26(0) - 20(1) + 19 = 1 - 20 + 19 = 0$.Thus,the circle $x^2+y^2-26x-20y+19=0$ passes through $(0, 1)$.

$A$ circle $C$ touches the $X$-axis and makes an intercept of length $2$ units on the $Y$-axis. If the centre of this circle lies on the line $y=x+1$,then which of the following is a circle passing through the centre of the circle $C$?

Similar Questions

$A$ circle touches both the coordinate axes and the straight line $L \equiv 4x+3y-6=0$ in the first quadrant. If this circle lies below the line $L=0$,then the equation of that circle is

Find the equation of the circumcircle of the triangle formed by the lines $y \pm \sqrt{3}x = 6$ and the $x$-axis.

Two concentric circles are given,where the equation of the smaller circle is $x^2 + y^2 = 4$. If each circle makes an intercept on the line $x + y = 2$ and the length of the intercept formed between the two circles is $1$,then the equation of the larger circle is:

Find the equation of the diameter of the circle $x^2 + y^2 - 4x + 2y - 11 = 0$ which corresponds to the system of parallel chords $x - 2y + c = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module