A circle has its centre in the first quadrant | vedclass

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

Question

(D) Let the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(2,3)$,we have $(2-h)^2 + (3-k)^2 = r^2$. The intercept on the line $x=2$ is

Vedclass · Accepted Answer

$x^2+y^2-8x-9y+30=0$

Vedclass · Answer

(D) Let the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(2,3)$,we have $(2-h)^2 + (3-k)^2 = r^2$. The intercept on the line $x=2$ is $2\sqrt{r^2 - (2-h)^2} = 3$. Substituting $r^2 - (2-h)^2 = (3-k)^2$,we get $2\sqrt{(3-k)^2} = 3$,so $|3-k| = 1.5$,which means $k = 4.5$ or $k = 1.5$. The intercept on the line $y=3$ is $2\sqrt{r^2 - (3-k)^2} = 4$. Substituting $r^2 - (3-k)^2 = (2-h)^2$,we get $2\sqrt{(2-h)^2} = 4$,so $|2-h| = 2$,which means $h = 4$ or $h = 0$. Since the centre $(h,k)$ is in the first quadrant,$h, k > 0$. Testing the case $(h,k) = (4, 4.5)$: $r^2 = (2-4)^2 + (3-4.5)^2 = 4 + 2.25 = 6.25$. The equation is $(x-4)^2 + (y-4.5)^2 = 6.25$,which simplifies to $x^2 - 8x + 16 + y^2 - 9y + 20.25 = 6.25$,or $x^2 + y^2 - 8x - 9y + 30 = 0$.

$A$ circle has its centre in the first quadrant and passes through $(2,3)$. If this circle makes intercepts of length $3$ and $4$ respectively on $x=2$ and $y=3$,its equation is

Similar Questions

The polar equation of the circle with centre $\left(2, \frac{\pi}{2}\right)$ and radius $3$ units is :

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

The equations of the circles touching both the axes and passing through the point $(1, 2)$ are

If the parametric equations of the circle passing through the points $(3,4)$,$(3,2)$,and $(1,4)$ are $x=a+r \cos \theta$ and $y=b+r \sin \theta$,then find the value of $b^{a} r^{a}$.

If the power of the point $(1, 6)$ with respect to the circle $x^2 + y^2 + 4x - 6y - a = 0$ is $-16$,then $a =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module