Q. There are 2 circles. 1 circle has its center as (0,5) and radius = 5 units. The other circle has its center as (12,0) and radius =12 units. A third circle is drawn which will pass through the center of the second circle and the two points of intersection of the other two circles. Find the radius of such a circle.

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TopNewestThe equation of the first circle is \(x^2 + (y-5)^2 = 25\)

The equation of the second circle is \((x-12)^2 + y^2 = 144\)

Solving them together, we get the two points of intersection of both the circles as \((0,0)\) and \(\left( \dfrac{600}{169}, \dfrac{1440}{169} \right) \).

Centre of the second circle is \((0,5)\)

So the centre of the third circle ( say \((h,k)\) ) must be equidistant from the three points \((0,0)\ ; \left( \dfrac{600}{169}, \dfrac{1440}{169} \right) \ ; (0,5)\)

Using distance formula, find the value of \(h\) and \(k\). Radius of the circle would be the \(\sqrt{h^2 + k^2}\).

Do it yourself: Finding the values of \(h,k\) – Satyajit Mohanty · 1 year, 7 months agoLog in to reply

@Calvin Lin @Chew-Seong Cheong @Satyajit Mohanty @Pi Han Goh @Nihar Mahajan and all others.pPlease help – Anik Mandal · 1 year, 7 months ago

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Is Answer 6 units – Hardik Gupta · 1 year, 7 months ago

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– Pramita Kastha · 1 year, 7 months ago

How? pls explainLog in to reply