Find the equation of the circle touching the line 2x + 3y + 1 =0 at the point (1, -1) and is orthogonal to the circle which has the line segment having end points (0, 3) and (-2, -1) as the diameter

Dear Student,
Please find below the solution to the asked query:

Diametric end of given circle are 0,3 and -2,-1.Hence equation of cirlce isx-0x+2+y-3y+1=0x2+2x+y2-2y-3=0x2+y2+2x-2y-3=0Let equation of required circle be x2+y2+2gx+2fy+c=0Applying condition of orthogonality we get:2g1+2f-1=c-32g-2f=c-3....i1,-1 lies on curve1+1+2g-2f+c=0 2g-2f=-c-2....iiBy i and iic-3=-c-2c=12i becomes2g-2f=12-3=-52g-f=-54...iiix2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+12=02x2+2y2+4gx+4fy+1=0Now as curve and line touch each other at 1,-1 hence both will havsame slope at that point.Slope of line 2x+3y+1=0 is -23Now2x2+2y2+4gx+4fy+1=0Differentiate with respect to x4x+4y.dydx+4g+4f.dydx=0x+y.dydx+g+f.dydx=0At 1,-1, dydx=-231-1-23+g+f-23=01+23+g-23f=0Use iii1+23+f-54-23f=0f3=54-53f3=-512f=-54Put in ii to getg=-52Hence equation of circle is2x2+2y2-10x-5y+1=0

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.

  • 18
What are you looking for?