Circles
6 Board Maths previous year questions on Circles — options free on every question; 1 include the answer & explanation free, the rest unlock with PYQ Pass.
Let the set of all values of \(r\), for which the circles \({(x+1)}^{2}+{(y+4)}^{2}={r}^{2}\) and \({x}^{2}+{y}^{2}−4x−2y−4=0\)
intersect at two distinct points be the interval \(\left(\alpha ,\beta \right)\). Then \(\alpha \beta\) is equal to
[JEE Main 2026, 22 Jan (Shift 1)]
\(25\)
Given circles:
\({(x−2)}^{2}+{(y−1)}^{2}={3}^{2}\\ \text{and}{(x+1)}^{2}+{(y+4)}^{2}={r}^{2}\)
Since the circles intersect each other at two different points.
So, \(\left|{r}_{1}−{r}_{2}\right|<{c}_{1}{c}_{2}<{r}_{1}+{r}_{2}\)
\(\Rightarrow \left|r−3\right|<\sqrt{{(2+1)}^{2}+{(1+4)}^{2}} \(\Rightarrow \left|r−3\right|<\sqrt{34}\text{ }\&\text{ }r+3>\sqrt{34}\) \(\Rightarrow −\sqrt{34} i.e. \(r=\left(3−\sqrt{34},\text{ }3+\sqrt{34}\right)\cap \left(\sqrt{34}−3,∞\right)\) i.e. \(r\in \left(\sqrt{34}−3,\sqrt{34}+3\right)\) \(∴\alpha \beta =\left(\sqrt{34}−3\right)\left(\sqrt{34}+3\right)\) \(=34–9=25\)
The tangents drawn at the ends of any diameter of a circle are mutually :
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If the circumference of a circle is 176 cm , then its radius is :
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If the volume of a sphere is \(36\pi c{m}^{3}\), then its radius is :
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If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of \(80^\circ\), then \(∠POA\) is equal to
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Radius of a circle is 3.5 cm. Find its circumference.
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