21.0 Family of circles

  Circles

1. Equation of family of circles passing through the point of intersection of two given circles $S=0$ and $S'=0$ is

$S + \lambda S' = 0$, where $\lambda $ is a parameter and $\lambda \in R - \{ - 1\} $.

2. Equation of family of circles passing through the point of intersection of two given circles

$S=0$ and a line $L=0$ is $S + \lambda L = 0$ where $\lambda $ is a parameter.

3. Equation of family of circles passing through two given points $P({x_1},{y_1})$ and $Q({x_2},{y_2})$ is

$$\left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) + \lambda \left| {\begin{array}{c}x&y&1 \\{{x_1}}&{{y_1}}&1 \\{{x_2}}&{{y_2}}&1\end{array}} \right| = 0$$

4. Family of circles circumscribing a triangle whose sides are given by ${L_1} = 0$, ${L_2} = 0$ and ${L_3} = 0$ is given by $${L_1}{L_2} + \lambda {L_2}{L_3} + \mu {L_3}{L_1} = 0$$

Provided coefficient of $xy=0$ and coefficient of ${x^2}$$=$coefficient of ${y^2}$ using which $\lambda $ and $\mu $ can be find out.

5. Equation of circle circumscribing a quadrilateral whose sides in order are given by ${L_1} = 0$, ${L_2} = 0$, ${L_3} = 0$ and ${L_4} = 0$ are $$\mu {L_1}{L_3} + \lambda {L_2}{L_4} = 0$$

provided coefficient of $xy=0$ and coefficient of ${x^2}$$=$coefficient of ${y^2}$ using which $\lambda $ and $\mu $ can be find out.

Question 29. Find the equation of circle passing through the intersection of two circles $S \equiv {x^2} + {y^2} + 13x - 3y = 0$ and $S' \equiv 2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and also passes through $(1,1)$.

Solution: Equation of family of circles passing through the point of intersection of two given circles $S=0$ and $S'=0$ is $S + \lambda S' = 0$.

$$({x^2} + {y^2} + 13x - 3y) + \lambda \left( {2{x^2} + 2{y^2} + 4x - 7y - 25} \right) = 0$$

The above equation passes through $(1,1)$, therefore $$({1^2} + {1^2} + 13 \times 1 - 3 \times 1) + \lambda \left( {2 \times {1^2} + 2 \times {1^2} + 4 \times 1 - 7 \times 1 - 25} \right) = 0$$

or, $$12 = \frac{{24}}{2}\lambda $$

Therefore, $$\lambda = 1$$

The equation of circle passing through the intersection of two circles is $$({x^2} + {y^2} + 13x - 3y) + 1\left( {2{x^2} + 2{y^2} + 4x - 7y - 25} \right) = 0$$ $$3{x^2} + 3{y^2} + 17x - 10y - 25 = 0$$

Question 30. Tangents are drawn from the point $P(1,8)$ to the circle $S \equiv {x^2} + {y^2} - 6x - 4y - 11 = 0$ which touches the circle at $A$ and $B$. Find equation of circle passing through points $P$, $A$ and $B$.

Solution: As from figure $56$, $AB$ is the chord of contact and its equation can be find out using $T=0$ i.e., $$x + 8y - 3\left( {x + 1} \right) - 2\left( {y + 8} \right) - 11 = 0$$ $$3y-x=15$$

Now, the equation of circle passing through $P$, $A$ and $B$ is converted to a circle passing through the intersection of a circle $S \equiv {x^2} + {y^2} - 6x - 4y - 11 = 0$ and a line $3y-x=15$.

Therefore, the equation can be written in the form of $S + \lambda L = 0$ i.e., $$({x^2} + {y^2} - 6x - 4y - 11) + \lambda \left( {3y - x - 15} \right) = 0$$

which satisfies the point $P(1,8)$ $$({1^2} + {8^2} - 6 \times 1 - 4 \times 8 - 11) + \lambda \left( {3 \times 8 - 1 - 15} \right) = 0$$ $$\lambda = - 2$$

Therefore, the equation of circle is $$({x^2} + {y^2} - 6x - 4y - 11) - 2\left( {3y - x - 15} \right) = 0$$ $${x^2} + {y^2} - 4x - 10y + 19 = 0$$

NOTE: In such type of case if the circle is passing through points $P$, $A$ and $B$ then it will also pass through the centre of the circle.