Circles
Circle
nta_pyq_2025_apr
Grade 11
Question:
A circle $C$ of radius $2$ lies in the second quadrant and touches both the coordinate axes. Let $r$ be the radius of a circle that has centre at the point $(2, 5)$ and intersects the circle $C$ at exactly two points. If the set of all possible values of $r$ is the interval $(\alpha, \beta)$, then $3\beta - 2\alpha$ is equal to
Step-by-Step Solution
Key Concept: Two circles intersect at exactly two points when $|r_1-r_2| < d < r_1+r_2$ (strict inequalities); compute the distance between the two centres and solve for the open interval of $r$.
Circle $C$ in the second quadrant touching both axes with radius $2$ has centre $C_1=(-2,2)$. Let $C_2=(2,5)$ with radius $r$. Distance $d=C_1C_2=\sqrt{(2+2)^2+(5-2)^2}=\sqrt{25}=5$. For exactly two intersections: $|r-2|<5$ and $5<r+2$, giving $r>3$ and $r<7$. So $\alpha=3$, $\beta=7$, and $3\beta-2\alpha=21-6=15$.
Correct Answer: 2