Question

Mark (a) if both A and R are correct and R is the correct explanation for A.
Mark (b) if both A and R are correct R is not the correct explanation for A.
Mark (c) if A is true but R is false
Mark (d) if A is false but R is true
A: The circle \[{{x}^{2}}+{{y}^{2}}+2ax+c=0,{{x}^{2}}+{{y}^{2}}+2by+c=0\] touch if \[\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}=\dfrac{1}{c}\].
R: Two circles with centers \[{{C}_{1}},{{C}_{2}}\] and radii \[{{r}_{1}},{{r}_{2}}\] touch each other if \[{{r}_{1}}\pm {{r}_{2}}={{C}_{1}}{{C}_{2}}\]

Answer 1

Hint: To choose the correct option, first of all, we will check whether statement R is true or not. If statement R is found true, then we have to check which of the option (a), (b), and (d) satisfies the condition and if R is false then the possible correct option is (c).

Complete step-by-step answer:
In this question, we have to check whether A and R are true or not, and if they are true, then we need to check if they have any relation between them. The statement R states that two circles with the centers \[{{C}_{1}}\] and \[{{C}_{2}}\] and radii \[{{r}_{1}}\] and \[{{r}_{2}}\] touch each other if \[{{r}_{1}}\pm {{r}_{2}}={{C}_{1}}{{C}_{2}}\]. Now, we will consider both the cases of two circles touching each other internally and externally.
When circles touch internally

Here, we can see that \[{{C}_{2}}P={{r}_{2}}\] is the radius of the circle with the center \[{{C}_{2}}\] and \[{{C}_{1}}P={{r}_{1}}\] is the radius of the circle with the center \[{{C}_{1}}\] and from the figure, we can see that,
\[{{C}_{2}}P-{{C}_{1}}P={{C}_{1}}{{C}_{2}}\]
\[{{r}_{2}}-{{r}_{1}}={{C}_{1}}{{C}_{2}}.....\left( i \right)\]
When circle touches externally

Here also, we can see that \[{{C}_{2}}P={{r}_{2}}\] and \[{{C}_{1}}P={{r}_{1}}\] are radii of the circle with center \[{{C}_{2}}\] and \[{{C}_{1}}\] respectively and from the figure, we can say that
\[{{C}_{2}}P+{{C}_{1}}P={{C}_{1}}{{C}_{2}}\]
\[{{r}_{2}}+{{r}_{1}}={{C}_{1}}{{C}_{2}}....\left( ii \right)\]
From equation (i) and (ii), we can conclude that if \[{{C}_{1}}{{C}_{2}}={{r}_{1}}\pm {{r}_{2}}\], then two circles with centers \[{{C}_{1}}\] and \[{{C}_{2}}\] and radii \[{{r}_{1}}\] and \[{{r}_{2}}\] touches each other.
Hence, R is a true statement. So, option (c) cannot be the correct answer.
Now, we will try to derive A using R to see which option is correct. We have been given a condition that \[{{x}^{2}}+{{y}^{2}}+2ax+c=0\] and \[{{x}^{2}}+{{y}^{2}}+2by+c=0\] touches if \[\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}=\dfrac{1}{c}\]. So, if we prove that if the circle touches each other, then \[\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}=\dfrac{1}{c}\], then we will get our required result.
As we have proved that if the circles touch each other, then \[{{C}_{1}}{{C}_{2}}={{r}_{1}}\pm {{r}_{2}}\]. So, let us consider circle 1 as \[{{x}^{2}}+{{y}^{2}}+2ax+c=0\] and circle 2 as \[{{x}^{2}}+{{y}^{2}}+2by+c=0\]. We know that the center of a circle of the equation \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] is given by (– g, – f) and radius as \[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\].
So, we can write the center of circle 1 as \[{{C}_{1}}\left( -a,0 \right)\] and of circle 2 as \[{{C}_{2}}\left( 0,-b \right)\]. Also, we can write the radius of circle 1 as \[{{r}_{1}}=\sqrt{{{a}^{2}}-c}\] and radius of circle 2 as \[{{r}_{2}}=\sqrt{{{b}^{2}}-c}\].
So, we can write the distance between the center \[{{C}_{1}}\] and \[{{C}_{2}}\] as \[\sqrt{{{\left( 0-\left( -a \right) \right)}^{2}}+{{\left( -b-0 \right)}^{2}}}=\sqrt{{{a}^{2}}+{{b}^{2}}}\]
Now, we will put the value of \[{{C}_{1}}{{C}_{2}}\], \[{{r}_{1}}\] and \[{{r}_{2}}\] in the relation of the statement R. So, we get,
\[\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{{{a}^{2}}-c}\pm \sqrt{{{b}^{2}}-c}\]
Now, we will square both sides of the equation. So, we will get,
\[{{a}^{2}}+{{b}^{2}}={{\left( \sqrt{{{a}^{2}}-c} \right)}^{2}}+{{\left( \sqrt{{{b}^{2}}-c} \right)}^{2}}\pm 2\left( \sqrt{{{a}^{2}}-c} \right)\left( \sqrt{{{b}^{2}}-c} \right)\]
\[{{a}^{2}}+{{b}^{2}}={{a}^{2}}-c+{{b}^{2}}-c\pm 2\left( \sqrt{{{a}^{2}}-c} \right)\left( \sqrt{{{b}^{2}}-c} \right)\]
\[2c=\pm 2\left( \sqrt{{{a}^{2}}-c} \right)\left( \sqrt{{{b}^{2}}-c} \right)\]
\[c=\pm \left( \sqrt{{{a}^{2}}-c} \right)\left( \sqrt{{{b}^{2}}-c} \right)\]
Now, again we will square both sides of the equation. So, we will get,
\[{{c}^{2}}={{\left[ \left( \sqrt{{{a}^{2}}-c} \right)\left( \sqrt{{{b}^{2}}-c} \right) \right]}^{2}}\]
\[{{c}^{2}}=\left( {{a}^{2}}-c \right)\left( {{b}^{2}}-c \right)\]
\[{{c}^{2}}={{a}^{2}}{{b}^{2}}-{{b}^{2}}c-{{a}^{2}}c+{{c}^{2}}\]
\[{{a}^{2}}c+{{b}^{2}}c={{a}^{2}}{{b}^{2}}\]
Now, we will divide the whole equation by \[{{a}^{2}}{{b}^{2}}c\]. So, we will get,
\[\dfrac{{{a}^{2}}c}{{{a}^{2}}{{b}^{2}}c}+\dfrac{{{b}^{2}}c}{{{a}^{2}}{{b}^{2}}c}=\dfrac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}{{b}^{2}}c}\]
\[\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{a}^{2}}}=\dfrac{1}{c}\]
\[\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}=\dfrac{1}{c}\]
Here, we have proved that if the circles \[{{x}^{2}}+{{y}^{2}}+2ax+c=0\], \[{{x}^{2}}+{{y}^{2}}+2by+c=0\] touches each other then \[\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}=\dfrac{1}{c}\] which is the same as statement A.
Therefore, we can say that statement A and statement R are true and so we have derived statement A using statement R. So, we can say statement R is the correct explanation of statement A.
Hence, option (a) is the correct answer.

Note: In this question, one might choose any of the options - (a), (b) or (d) as the correct answer randomly without verifying statement A, by just after verifying the statement R as a true statement to either save time or reduce efforts which is not a correct way to solve a question. A better way to solve this is by patiently verifying statement A first by using statement R if satisfied then we will choose option (a) as correct otherwise we will look for another option and then choose the correct answer.Students should remember the center of a circle of the equation \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] is given by $(– g, – f)$ and radius as \[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\] and also distance between two points formula i.e \[\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\] for solving these types of questions.