The circles $\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}+16=0$ and $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$…
The circles $\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}+16=0$ and $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ intersect at two distinct points if
- $\mathrm{a} < 2$
- $2 < $ a $ < 8$
- $a>8$
- $\mathbf{a}=2$
Solution
For \(C_1: x^2+y^2 \rightarrow 10 x+16=0\); Centre \(=(5,0)\), radius \(=\sqrt{5^2+0-16}=3\)
For \(C_2: x^2+y^2=r^2 ;\) Centre \(=(0, 0)\), radius \(=r\)
For intersection, \(\left|\mathrm{r}_1-\mathrm{r}_1\right| < \mathrm{C}_1 \mathrm{C}_2\)
\(\Rightarrow \mathrm{r}-3 < 5 \Rightarrow \mathrm{r} < 8\)
and \(\mathrm{r}_1+\mathrm{r}_2 > \mathrm{C}_1 \mathrm{C}_2, \mathrm{r}+3 > 5 \Rightarrow \mathrm{r}=2\)
From (1) and (2), \(2 < \mathrm{r} < 8\).
Asked in: TEST SERIES MHT-CET Full Test 6
Practice more Circle questions on Aicharya