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.

Note by Pramita Kastha
2 years, 11 months ago

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The 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 $$h,k$$

- 2 years, 11 months ago

- 2 years, 11 months ago

- 2 years, 11 months ago

How? pls explain

- 2 years, 11 months ago