Geometry

Straight Lines

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The maximum distance between the point $(-5, 0)$ and a point on the circle $x^2 + y^2 = 4$ is

Entered answer:

✅ Correct Answer: 7

Solution figure for IPMAT Indore 2019 SA question 9 (Geometry)

We have the circle $x^2+y^2=4$ and the fixed point $P=(-5,0).$

We need to find the maximum distance between $P$ and any point on the circle.

The equation $x^2+y^2=4$ can be written as $x^2+y^2=2^2.$

Therefore, the circle has:

Centre: $(0,0)$
Radius: $2$

The fixed point is $P=(-5,0)$ and the centre is $O=(0,0).$

Therefore, $OP=\sqrt{(-5-0)^2+(0-0)^2}$ $=\sqrt{25}$ $=5.$

The maximum distance from a fixed point outside a circle to a point on the circle occurs when we move from the fixed point through the centre to the opposite end of the circle.

So, $\text{Maximum distance}=OP+r$

where $r=2$ .

Thus, $\text{Maximum distance}=5+2$ $={7}$

Answer: 7

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IPMAT Indore Past Year Questions (Topic-Wise)

The maximum distance between the point $(-5, 0)$ and a point on the circle $x^2 + y^2 = 4$ is