The equation of the circle, having the lines x^2+2 x y+ 3 x+6 y=0 as its normals and having size

The equation of the circle, having the lines x^2+2 x y+ 3...

Key Concepts

Circle Equation in Analytic Geometry

This concept involves expressing a circle in the coordinate plane by an equation usually in the form (x - h)² + (y - k)² = r², where (h, k) represents the center and r is the radius. Mastery of this form is fundamental as it allows one to analyze geometric properties, perform algebraic manipulation, and establish conditions like tangency and containment with other curves.

Normals and Their Properties

Normals to a curve, particularly a circle, are lines that are perpendicular to the tangent at the point of contact. In circles, a key property is that the normal at any point passes through the center of the circle. This relationship is crucial for determining the position of the center when given potential normals, as any line known to be a normal must contain the center.

Finding the Center of a Circle Using Normals

Since normals of a circle pass through its center, if one is given one or more lines that are guaranteed to be normals, the center must lie on each of these lines. In problems where multiple normal lines are provided, their intersection locates the center of the circle. This method reduces the problem of finding the center to solving a system of linear equations.

Enclosure of One Circle by Another

When one circle is required to fully contain another, the condition of tangency is often invoked. The minimal enclosing circle will have a radius that is exactly the sum of the radius of the contained circle and the distance between the centers. This concept of tangency (either internal or external) underpins many problems in circle geometry where one is tasked with ensuring that one geometric object just contains another.

Transcript

00:01 Hello everyone in this question it is said that the equation of circle having lines have lines equation of lines are x square plus 2 x y plus 3 x plus 6y is equal to 0 as its normal lines are as its normal and having size just sufficient to the circle sufficient to contain the circle x x minus 4 plus y y minus 3 is equals to 0 so we have to find the equation of circle first of all equation of line pair of lines pair of lines is the equation of line is given which is x square plus 2 x y plus 3 x plus 6 y plus 6 y is equal to 0.

01:03 Take common x and inside we get x plus 2y.

01:08 Take common 3 and we get also x plus 2 y is equal to 0.

01:13 So there are two lines x plus 2 y and x plus 3 is equal to 0 which is x plus 2 y is equal to 0 and x plus 3 is equal to 0 are two normals.

01:30 2 normals and their point of intersections are, point of intersection are.

01:45 So we have to find the point of intersection, which is x is equal to negative of 3, put value of x in this, minus 3 plus 2, y is equal to 0, so y becomes equal to 2, sorry 3 divide by 2.

02:01 So our center becomes minus 3 and 3 divide by 2 is the center of required circle.

02:12 Center of required circle.

02:19 Now, it is given the equation of circle, one circle is given which is x.

02:24 X minus 4 plus y, y minus 3 is equals to 0.

02:30 It is the diametric form of the circle is the diametric form of circle.

02:42 Hence their diametric end points are diametric end points are 0 and 4 .3.

02:57 These are the two diametric end points which means center is, center of the this circle is 0 plus 4 by using the midpoint formula 0 plus 3 divide by 2.

03:10 So center is 2 and 3 divide by 2.

03:14 Now we have to find the radius.

03:17 So radius of diametric circle becomes equals to square root of 2 2 2s are 4 plus 3 3 3 are 9 divided by 4.

03:26 Solve this we get 16 plus 9 divide by 4 which is equals to 25 divide by 4.

03:32 Which is equals to 5 divide by 2.

03:38 Now the required circle with center, we find the required circle with center with center minus 3 and 3 divide by 2...

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