A circle whose radius is $$5$$ and which touches externally the circle $$x2$$ $$+$$ $$y2$$ – $$2x$$ – $$4y$$ – $$20$$ $$=$$ $$0$$ at the point (5$$, $$5) intersects in real distinct points the line

Question

A circle whose radius is $$5$$ and which touches externally the circle $$x2$$ $$+$$ $$y2$$ – $$2x$$ – $$4y$$ – $$20$$ $$=$$ $$0$$ at the point (5$$, $$5) intersects in real distinct points the line

Infinity Learn · Accepted Answer

Centre of the given circle is A (1$$, $$2) and its radius is $$1+(2)2+20=5$$. Point of contact P is (5$$, $$5). Let B (h$$, $$k) be the centre of the required circle of radius $$5$$, then P is the $$mid-point$$ of AB, so $$thath+12=5$$  and   $$k+22=5$$⇒$$h=9$$,$$k=8and$$ an equation of the required circle is     $$x2+y2$$−$$18x$$−$$16y+120=0If$$ $$x=0$$, $$y2$$−$$16y+120=0$$ does not give real values of yIf $$y=0$$,$$x2$$−$$18x+120=0$$ does not give real values of xIf          $$y=x$$,$$2x2$$−$$34x+120=0or$$   $$x2$$−$$17x+60=0$$  ⇒     $$x=5$$,$$12This$$ shows that the circle intersects the line y $$=$$ x at two real distinct points.

A circle whose radius is $5$ and which touches externally the circle $x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ at the point $(5, 5)$ intersects in real distinct points the line

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A circle whose radius is 5 and which touches externally the circle x2 + y2 – 2x – 4y – 20 = 0 at the point (5, 5) intersects in real distinct points the line

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A circle whose radius is $5$ and which touches externally the circle $x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ at the point $(5, 5)$ intersects in real distinct points the line