Solveeit Logo
Login

Question

Question: The locus of centres of all circles which touch the line \[x = 2a\] and cut the circle \( {x^2} + {y...

The locus of centres of all circles which touch the line x=2ax = 2a and cut the circle x2+y2=a2{x^2} + {y^2} = {a^2} orthogonally is

Explanation

Solution

Hint : In this question, we are given the equation of a line and a circle and we have to find the locus of centres of all circles which touch that line and cut the given circle orthogonally, that is, we have to find the equation that is satisfied by the centre of all the circles following above-mentioned condition.

Complete step by step solution:
Let the coordinates of the centre of the circle that touches the line x=2ax = 2a and cut the circle x2+y2=a2{x^2} + {y^2} = {a^2} orthogonally be (h,k)(h,k)
As this circle cuts the line x=2ax = 2a and has its centre at (h,k)(h,k) , so the radius of the circle will be equal to the difference in the distance of the given line from the origin and the distance of the center of the given circle from the origin, so radius =h−2a= h - 2a
So, the equation of this circle will be –
(x−h)2+(y−k)2=r2 ⇒(x−h)2+(y−k)2=(h−2a)2 ⇒x2+h2−2xh+y2+k2−2ky=h2+4a2−4ah ⇒x2+y2−2hx−2ky+k2−4a2+4ah=0   {(x - h)^2} + {(y - k)^2} = {r^2} \\\ \Rightarrow {(x - h)^2} + {(y - k)^2} = {(h - 2a)^2} \\\ \Rightarrow {x^2} + {h^2} - 2xh + {y^2} + {k^2} - 2ky = {h^2} + 4{a^2} - 4ah \\\ \Rightarrow {x^2} + {y^2} - 2hx - 2ky + {k^2} - 4{a^2} + 4ah = 0 \;
We know that for two circles x2+y2+2g1x+2f1y+c1=0{x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0 and x2+y2+2g2x+2f2y+c2=0{x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0
to be orthogonal –
2g1g2+2f1f2=c1+c22{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}
Here g1=−h, g2=0, f1=−k, f2=0, c1=k2−4a2+4ah, c2=−a2{g_1} = - h,\,{g_2} = 0,\,{f_1} = - k,\,{f_2} = 0,\,{c_1} = {k^2} - 4{a^2} + 4ah,\,{c_2} = - {a^2}
⇒2(−h)0+2(−k)0=k2−4a2+4ah−a2 ⇒k2+4ah−5a2=0   \Rightarrow 2( - h)0 + 2( - k)0 = {k^2} - 4{a^2} + 4ah - {a^2} \\\ \Rightarrow {k^2} + 4ah - 5{a^2} = 0 \;
Now, for finding the general equation of the circles that touch the line x=2ax = 2a and cut the circle x2+y2=a2{x^2} + {y^2} = {a^2} orthogonally, we take k=yk = y and h=xh = x
⇒y2+4ax−5a2=0\Rightarrow {y^2} + 4ax - 5{a^2} = 0
Hence the locus of centres of all circles which touch the line x=2ax = 2a and cut the circle x2+y2=a2{x^2} + {y^2} = {a^2} orthogonally is y2+4ax−5a2=0{y^2} + 4ax - 5{a^2} = 0 .

Note : Two circles are said to be orthogonal if they intersect at two points such that the radii of both the circles drawn to the point of intersection meet at right angles. Thus the circle whose radius is perpendicular to the radius of another circle at two points is said to be perpendicular to the other circle.