Given the centre (0,0) and a tangent that has the equation 3x-4y=30, what is the equation of the circle?

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Answer 1 · 2016-07-17T07:58:16

#x^2+y^2=36#

Length of a perpendicular from a point #(x_1,y_1)# to the line #ax+by+c=0# is given by #(|ax_1+by_1+c|) /(sqrt(a^2+b^2))#

Hence length of perpendicular from #(0,0)# to #3x-4y-30=0# is #(|3*0+(-4)*0-30|) /(sqrt(3^2+(-4)^2))=30/(sqrt(9+16))=30/5=6#

But this is radius of the circle. Hence, equation of circle will be

#(x-0)^2+(y-0)^2=6^2# or

#x^2+y^2=36#

Answer 2 · 2016-07-17T08:06:12

Reqd. eqn. of the circle is #x^2+y^2=36#.

To determine the eqn. of a circle, we need its (i) centre, and, (ii) the

radius. As the centre is given to be the Origin #O(0,0)#, let us

suppose that the radius of the circle is #r#.

Now the line #l : 3x-4y-30=0# is tgt. to the circle. So, by

Geometry , we know that the **#bot# dist., say #d#, from the

Centre #O# to the tgt. line #l# must equal radius #r#.**

#:. d=r rArr |3(0)-4(0)-30|/sqrt{(3)^2+(4)^2}=r rArr r=30/5=6#

Now, the eqn. of a circle having centre at #(h,k)# & radius #r#, is

given by, #(x-h)^2+(y-k)^2=r^2#. Hence, in our case, the reqd.

eqn. is #x^2+y^2=36#.

Note : The #bot#dist. #d# from a pt. #(x_1,y_1)# to a line # l : ax+by+c=0#

is #d=|ax_1+by_1+c|/sqrt(a^2+b^2).#