A circle touches both the x-axis and the line 4 x-3 y+ 4=0 . If its centre is in the third quadr

A circle touches both the x-axis and the line 4 x-3 y+ 4=0 ....

Key Concepts

Distance from a Point to a Line

This concept uses the formula for the shortest distance between a point (h, k) and a line given by Ax + By + C = 0, which is |Ah + Bk + C|/?(A² + B²). It is essential when enforcing a tangency condition since a circle is tangent to a line if the distance from its center to the line equals its radius.

Circle Tangency Condition

A circle is defined by its center (h, k) and radius r, expressed by the equation (x - h)² + (y - k)² = r². For a line to be tangent to the circle, the perpendicular distance from the center to the line must equal the circle’s radius. This condition creates a crucial relationship used to determine unknown parameters in problems involving tangency.

Coordinate Geometry Constraints and Quadrant Conditions

In coordinate geometry, additional constraints such as the center of a circle lying on a specific line or within a particular quadrant help limit the possible values of the center coordinates. For example, specifying that the center is in the third quadrant implies that both h and k are negative, which critically steers the solution space when combined with other geometric conditions.

System of Equations in Coordinate Geometry

By combining multiple geometric conditions—such as tangency to lines and the center lying on a given line—a system of equations is formed. Solving this system allows the determination of unknown parameters like the center coordinates and the radius, thereby fully characterizing the circle. This approach illustrates the interplay between algebra and geometry in problem solving.

Transcript

00:01 In this problem we have been given a function of 0 to infinity to r satisfies the equation f of xy is equal to 2 into f of x minus f of x by x by y.

00:09 If f is differentiable on r and f of 1 is equal to 0, f dash of 1 equal to 1 then so let's proceed.

00:18 We have so f of x comma y is equal to 2 into f of x minus f of x by y.

00:31 So, putting x equal to 1, we would get 2 into f of 1 is equal to f of y minus 1 by y.

00:44 We would get therefore f of y is equal to minus f of 1 by y since f of 1 is equal to 0 as given.

01:00 Therefore f dash x now would be equal to limit h tends to 0 f of x plus h minus f of x upon h yeah so now therefore we would get this is equal to limit h tends to 0 f of x into 1 plus h by x minus f of x of x upon h upon h so this would be equal to limit extends to 0 2x x into f of 1 plus h by x minus f of 1 upon h by x minus 1 by x minus 1 by x square into f of 1 by x uh yeah so f of 1 by x plus 1 by x plus h by x squared plus f of x upon h by x square so this would give us therefore this would be use equation two so therefore we will get we got the previous step by using the property f of y is minus f of one by y so using that property and now we got this is equal to 2x minus 1x square into f dash of 1 by x so differentiating equation 2 which is f of y is equal to minus f of 1 by y this was equation number 2 which we had previously got so differentiating it with respect to y we would get f dash of y would be equal to f dash of 1 by y into 1 by y squared so we would get therefore f dash x is equal to 2 by x minus f dash x that implies f dash x equal to 1 by x so we would get therefore d f f is equal to 1 by x d x d x is equal to 1 by x d x on integrating, we would get f of x equal to lnx plus c...

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