Equation of a circle is (x − a)^2 + (y − b)^2 = r^2. A point (b-a, a-b) is outside the circle. Find the equation of the tangent to the circle.
JEE nostalgia with this problem 
But let me just tell you the steps of how I would solve it.
- Center of the circle is at (a,b) from the given equation. Say O is the center
- Let’s consider point A and B to be the points on the circle where the tangent touches the circle.There are always two points on the circle where the tangent touches the circle if the point from which the lines are being drawn is outside the circle. So OA = OB = r
- The external point is say T (b-a, a-b).
So now consider a triangle OAT. you know A is a right angle. You know the distance lengths OA and OT. So basically you can calculate angle OTA. - The slope of the line TO is known (since we know the coordinates) and hence angle of that line from x axis is known. So slope of line TA can be obtained by adding the angle OTA to the angle of the line TO and getting the tan of it.
- For slope of the line TB, instead of adding the angle OTA, subtract it from the line TO’s angle.
- Write the equation of a line from the coordinates of the point T and the slope we just obtained.
Alternate method (Not sure if this gives 2 distinct equations in x and y)
- Consider point A to have coordinates (x,y). Follow till step 3 above and apply the formula that product of slopes of TA and OA should be -1 (Since they are perpendicular)
Or let’s just wait for somebody else to answer 
Thank u very much. the five options will be equations in y=mx+b form, from which we can get the slope. Then how do find which is perpendicular if don’t the point A(x,y) in the alternate method.
In that case, we can use the formula for “ditance of a point from a line”. We know the line equation, and we know the point (center of the circle). We can verify if the distance we are getting is equal to r^2 from the options given
thanks, but for gre do we need to know the formula for ditance of a point from a line, can it be solved in any other way