If the circles x^2+y^2-16x-20y+164=r^2 (r0) | vedclass

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

Question

(B) The given circles are $S_1: x^2+y^2-16x-20y+164-r^2=0$ and $S_2: x^2+y^2-8x-14y+29=0$. For $S_1$,the center $C_1 = (8, 10)$ and radius $r_1 = \sqr

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The given circles are $S_1: x^2+y^2-16x-20y+164-r^2=0$ and $S_2: x^2+y^2-8x-14y+29=0$. For $S_1$,the center $C_1 = (8, 10)$ and radius $r_1 = \sqrt{8^2+10^2-(164-r^2)} = \sqrt{64+100-164+r^2} = \sqrt{r^2} = r$ (since $r>0$). For $S_2$,the center $C_2 = (4, 7)$ and radius $r_2 = \sqrt{4^2+7^2-29} = \sqrt{16+49-29} = \sqrt{36} = 6$. The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(8-4)^2+(10-7)^2} = \sqrt{4^2+3^2} = \sqrt{16+9} = 5$. Since the circles intersect at two distinct points,the condition is $|r_1-r_2| Substituting the values,we get $|r-6| From $5  -1$. Since $r>0$,this implies $r > 0$. From $|r-6| Combining these,$r \in (1, 11)$. The maximum integral value of $r$ is $10$.

If the circles $x^2+y^2-16x-20y+164=r^2$ $(r>0)$ and $x^2+y^2-8x-14y+29=0$ intersect in two distinct points,then the maximum possible integral value of $r$ is

Similar Questions

If the circles given by $S \equiv x^2+y^2-14x+6y+33=0$ and $S^{\prime} \equiv x^2+y^2-a^2=0$ $(a \in N)$ have $4$ common tangents,then the possible number of circles $S^{\prime}=0$ is

Two tangents are drawn from a point $P$ on the radical axis to two circles,touching them at $Q$ and $R$ respectively. Then,the triangle formed by joining $P, Q,$ and $R$ is:

The angle of intersection of the circles $x^2 + y^2 - x + y - 8 = 0$ and $x^2 + y^2 + 2x + 2y - 11 = 0$ is

The co-axial system of circles given by ${x^2} + {y^2} + 2gx + c = 0$ for $c < 0$ represents

Find the value of $m+n$,if the circumference of the circle $x^2+y^2+8x+8y-m=0$ is bisected by the circle $x^2+y^2-2x+4y+n=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module