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If the circles ${x^2} + {y^2} + 6x + 8y + 16 = 0$ and ${x^2} + {y^2} + 2\\left( {3 - \\sqrt 3 } \\right)x + 2\\left( {4 - \\sqrt 6 } \\right)y = k + 6\\sq...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/if-the-circles-x2-y2-6x-8y-16-0-and-x2-y2-2-3-3-x-2-4-6-y-k-6-3-8-6-k-an"}}}]}20:T53e,

The circle $${x^2} + {y^2} + 6x + 8y + 16 = 0$$ has centre $$( - 3, - 4)$$ and radius 3 units.

The circle $${x^2} + {y^2} + 2\left( {3 - \sqrt 3 } \right)x + 2\left( {4 - \sqrt 6 } \right)y = k + 6\sqrt 3 + 8\sqrt 6 ,\,k > 0$$ has centre $$\left( {\sqrt 3 - 3,\,\sqrt 6 - 4} \right)$$ and radius $$\sqrt {k + 34} $$

$$\because$$ These two circles touch internally hence

$$\sqrt {3 + 6} = \left| {\sqrt {k + 34} - 3} \right|$$

Here, $$k = 2$$ is only possible ($$\because$$ $$k > 0$$)

Equation of common tangent to two circles is $$2\sqrt 3 x + 2\sqrt 6 y + 16 + 6\sqrt 3 + 8\sqrt 6 + k = 0$$

$$\because$$ $$k = 2$$ then equation is

$$x + \sqrt 2 y + 3 + 4\sqrt 2 + 3\sqrt 3 = 0$$ ...... (i)

$$\because$$ ($$\alpha$$, $$\beta$$) are foot of perpendicular from $$( - 3, - 4)$$

To line (i) then

$${{\alpha + 3} \over 1} = {{\beta + 4} \over {\sqrt 2 }} = {{ - \left( { - 3 - 4\sqrt 2 + 3 + 4\sqrt 2 + 3\sqrt 3 } \right)} \over {1 + 2}}$$

$$\therefore$$ $$\alpha + 3 = {{\beta + 4} \over {\sqrt 2 }} = - \sqrt 3 $$

$$ \Rightarrow {\left( {\alpha + \sqrt 3 } \right)^2} = 9$$ and $${\left( {\beta + \sqrt 6 } \right)^2} = 16$$

$$\therefore$$ $${\left( {\alpha + \sqrt 3 } \right)^2} + {\left( {\beta + \sqrt 6 } \right)^2} = 25$$